A force of magnitude 50 N is acting at an angle of 30° on a 2 kg object to push the object against a rough horizontal ceiling as shown in the diagram below. The object accelerates to the LEFT at 20 m.s-2. Ignore air resistance.

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Asked on September 11, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-09-11T18:51:29.551Z

Here's a step-by-step solution to the problem: 2.1 State Newton's second law of motion in words. Newton's second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The direction of the acceleration is the same as the direction of the net force. 2.2 Draw a labelled free-body diagram indicating all the forces acting on the object while in motion. Assumption: Based on the text "push the object against a rough horizontal ceiling" and "accelerates to the LEFT", the applied force of 50 N at 30^ is directed upwards and to the left, 30^ above the horizontal. The diagram's arrow direction is interpreted as indicating the angle and magnitude, but the actual direction of the force is consistent with the motion and interaction with the ceiling. [scale=0.8] % Object [draw, rectangle, minimum width=1.5cm, minimum height=1cm] (object) at (0,0) ; at (0,0) 2 kg; % Center of object (center) at (0,0); % Forces % Weight (Fg) [->, thick] (center) -- (0,-2) node[below] F_g; % Normal Force (Fn) - from ceiling, downwards [->, thick] (center) -- (0,-3.5) node[below right] F_N; % Applied Force (F) - up and left, 30 deg above horizontal [->, thick] (center) -- (-3cos30, 3sin30) node[above left] F = 50 N; [dashed] (center) -- (-3,0); % Horizontal reference [draw, angle radius=0.5cm, "30^", angle eccentricity=1.2] angle = (-3,0)--(center)--(-3cos30, 3sin30); % Kinetic Friction (Fk) - right (opposing leftward motion) [->, thick] (center) -- (2,0) node[right] F_k; % Acceleration (a) - left [->, ultra thick, blue] (center) -- (-2.5,0) node[above] a = 20 m/s^2; % Ceiling [thick] (-4,1.5) -- (4,1.5); in -3.5,-2.5,...,3.5 (,1.5) -- (+0.5,1.7); 2.3 Calculate the: 2.3.1 Magnitude of the normal force acting on the object. Step 1: Identify the forces in the vertical direction. The object is not accelerating vertically, so the net force in the vertical direction is zero ( F_y = 0). The forces acting vertically are: • Weight (F_g) acting downwards. • Normal force (F_N) from the ceiling, acting downwards (as the object is pushed against the ceiling from below). • Vertical component of the applied force (F_y) acting upwards. Step 2: Calculate the weight of the object. F_g = mg F_g = (2 kg)(9.8 m/s^2) F_g = 19.6 N Step 3: Calculate the vertical component of the applied force. F_y = F () F_y = (50 N) (30^) F_y = (50 N)(0.5) F_y = 25 N Step 4: Apply Newton's second law in the vertical direction. Taking upwards as positive: F_y = F_y - F_g - F_N = 0 25 N - 19.6 N - F_N = 0 5.4 N - F_N = 0 F_N = 5.4 N The magnitude of the normal force is 5.4 N. 2.3.2 Magnitude of the coefficient of kinetic friction. Step 1: Identify the forces in the horizontal direction. The object accelerates to the left, so the net force in the horizontal direction is F_x = ma_x. The forces acting horizontally are: • Horizontal component of the applied force (F_x) acting to the left. • Kinetic friction (F_k) acting to the right (opposing the leftward motion). Step 2: Calculate the horizontal component of the applied force. F_x = F () F_x = (50 N) (30^) F_x = (50 N)(sqrt(3)2) F_x ≈ 43.30 N Step 3: Apply Newton's second law in the horizontal direction. Taking leftwards as positive: F_x = F_x - F_k = ma_x We know F_k = _k F_N. F_x - _k F_N = ma_x (43.30 N) - _k (5.4 N) = (2 kg)(20 m/s^2) 43.30 N - 5.4 _k = 40 N Step 4: Solve for the coefficient of kinetic friction (_k). 43.30 - 40 = 5.4 _k 3.30 = 5.4 _k _k = (3.30)/(5.4) _k ≈ 0.611 The magnitude of the coefficient of kinetic friction is 0.61. 2.4 The force is now acting at a smaller angle. What effect will this have on the following? Choose from INCREASES, DECREASES or REMAINS THE SAME. 2.4.1 Value calculated in QUESTION 2.3.2 The value calculated in QUESTION 2.3.2 is the coefficient of kinetic friction (_k). The coefficient of kinetic friction depends only on the nature of the surfaces in contact, not on the angle or magnitude of the applied force. The value calculated in QUESTION 2.3.2 REMAINS THE SAME. 2.4.2 Kinetic frictional force acting on the object. Give a reason for the answer. The kinetic frictional force is given by F_k = _k F_N. From 2.3.1, the normal force is F_N = F () - F_g. If the angle (above the horizontal) becomes smaller, then () will decrease. Since F_N = F () - F_g, a decrease in F () will cause F_N to decrease. As _k remains the same (from 2.4.1) and F_N decreases, the kinetic frictional force F_k = _k F_N will DECREASE. Reason: A smaller angle of the applied force (above the horizontal) means a smaller upward vertical component of the applied force. This results in a smaller normal force exerted by the ceiling on the object, which in turn leads to a smaller kinetic frictional force. Send me the next one 📸

Accepted Answer · 2026-09-11T18:51:29.551Z

Here's a step-by-step solution to the problem: 2.1 State Newton's second law of motion in words. Newton's second law states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The direction of the acceleration is the same as the direction of the net force. 2.2 Draw a labelled free-body diagram indicating all the forces acting on the object while in motion. Assumption: Based on the text "push the object against a rough horizontal ceiling" and "accelerates to the LEFT", the applied force of 50 N at 30^ is directed upwards and to the left, 30^ above the horizontal. The diagram's arrow direction is interpreted as indicating the angle and magnitude, but the actual direction of the force is consistent with the motion and interaction with the ceiling. [scale=0.8] % Object [draw, rectangle, minimum width=1.5cm, minimum height=1cm] (object) at (0,0) ; at (0,0) 2 kg; % Center of object (center) at (0,0); % Forces % Weight (Fg) [->, thick] (center) -- (0,-2) node[below] F_g; % Normal Force (Fn) - from ceiling, downwards [->, thick] (center) -- (0,-3.5) node[below right] F_N; % Applied Force (F) - up and left, 30 deg above horizontal [->, thick] (center) -- (-3cos30, 3sin30) node[above left] F = 50 N; [dashed] (center) -- (-3,0); % Horizontal reference [draw, angle radius=0.5cm, "30^", angle eccentricity=1.2] angle = (-3,0)--(center)--(-3cos30, 3sin30); % Kinetic Friction (Fk) - right (opposing leftward motion) [->, thick] (center) -- (2,0) node[right] F_k; % Acceleration (a) - left [->, ultra thick, blue] (center) -- (-2.5,0) node[above] a = 20 m/s^2; % Ceiling [thick] (-4,1.5) -- (4,1.5); in -3.5,-2.5,...,3.5 (,1.5) -- (+0.5,1.7); 2.3 Calculate the: 2.3.1 Magnitude of the normal force acting on the object. Step 1: Identify the forces in the vertical direction. The object is not accelerating vertically, so the net force in the vertical direction is zero ( F_y = 0). The forces acting vertically are: • Weight (F_g) acting downwards. • Normal force (F_N) from the ceiling, acting downwards (as the object is pushed against the ceiling from below). • Vertical component of the applied force (F_y) acting upwards. Step 2: Calculate the weight of the object. F_g = mg F_g = (2 kg)(9.8 m/s^2) F_g = 19.6 N Step 3: Calculate the vertical component of the applied force. F_y = F () F_y = (50 N) (30^) F_y = (50 N)(0.5) F_y = 25 N Step 4: Apply Newton's second law in the vertical direction. Taking upwards as positive: F_y = F_y - F_g - F_N = 0 25 N - 19.6 N - F_N = 0 5.4 N - F_N = 0 F_N = 5.4 N The magnitude of the normal force is 5.4 N. 2.3.2 Magnitude of the coefficient of kinetic friction. Step 1: Identify the forces in the horizontal direction. The object accelerates to the left, so the net force in the horizontal direction is F_x = ma_x. The forces acting horizontally are: • Horizontal component of the applied force (F_x) acting to the left. • Kinetic friction (F_k) acting to the right (opposing the leftward motion). Step 2: Calculate the horizontal component of the applied force. F_x = F () F_x = (50 N) (30^) F_x = (50 N)(sqrt(3)2) F_x ≈ 43.30 N Step 3: Apply Newton's second law in the horizontal direction. Taking leftwards as positive: F_x = F_x - F_k = ma_x We know F_k = _k F_N. F_x - _k F_N = ma_x (43.30 N) - _k (5.4 N) = (2 kg)(20 m/s^2) 43.30 N - 5.4 _k = 40 N Step 4: Solve for the coefficient of kinetic friction (_k). 43.30 - 40 = 5.4 _k 3.30 = 5.4 _k _k = (3.30)/(5.4) _k ≈ 0.611 The magnitude of the coefficient of kinetic friction is 0.61. 2.4 The force is now acting at a smaller angle. What effect will this have on the following? Choose from INCREASES, DECREASES or REMAINS THE SAME. 2.4.1 Value calculated in QUESTION 2.3.2 The value calculated in QUESTION 2.3.2 is the coefficient of kinetic friction (_k). The coefficient of kinetic friction depends only on the nature of the surfaces in contact, not on the angle or magnitude of the applied force. The value calculated in QUESTION 2.3.2 REMAINS THE SAME. 2.4.2 Kinetic frictional force acting on the object. Give a reason for the answer. The kinetic frictional force is given by F_k = _k F_N. From 2.3.1, the normal force is F_N = F () - F_g. If the angle (above the horizontal) becomes smaller, then () will decrease. Since F_N = F () - F_g, a decrease in F () will cause F_N to decrease. As _k remains the same (from 2.4.1) and F_N decreases, the kinetic frictional force F_k = _k F_N will DECREASE. Reason: A smaller angle of the applied force (above the horizontal) means a smaller upward vertical component of the applied force. This results in a smaller normal force exerted by the ceiling on the object, which in turn leads to a smaller kinetic frictional force. Send me the next one 📸

A force of magnitude 50 N is acting at an angle of 30° on a 2 kg object to push the object against a rough horizontal ceiling as shown in the diagram below. The object accelerates to the LEFT at 20 m.s-2. Ignore air resistance.

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A force of magnitude 50 N is acting at an angle of 30° on a 2 kg object to push the object against a rough horizontal ceiling as shown in the diagram below. The object accelerates to the LEFT at 20 m.s-2. Ignore air resistance.

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