An object of mass 24 kg is on a frictionless inclined plane at an angle of 37 degrees. If it starts from rest and covers 18 m in 3.0 s, what is the applied force?

Here's a step-by-step solution to the problem:

Given Information:

• Mass of the object, $m = 24 \, \text{kg}$
• Angle of inclination, $\theta = 37^\circ$
• Initial velocity, $v_0 = 0 \, \text{m/s}$ (starts from rest)
• Distance covered, $d = 18 \, \text{m}$
• Time taken, $t = 3.0 \, \text{s}$
• Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$
• The plane is frictionless.

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Step 1: Calculate the acceleration ($a$) of the object.
Since the object starts from rest and moves a certain distance in a given time, we can use the kinematic equation:

$$d = v_0 t + \frac{1}{2}at^2$$

Given $v_0 = 0$:

$$d = \frac{1}{2}at^2$$ $$18 \, \text{m} = \frac{1}{2} a (3.0 \, \text{s})^2$$ $$18 = \frac{1}{2} a (9.0)$$ $$18 = 4.5a$$ $$a = \frac{18}{4.5}$$ $$a = 4.0 \, \text{m/s}^2$$

Step 2: Calculate the applied force ($F_{applied}$) acting on the object.
The forces acting along the incline are the applied force ($F_{applied}$) upwards and the component of gravity ($mg \sin\theta$) downwards. According to Newton's second law, the net force is $F_{net} = ma$.

$$F_{net} = F_{applied} - mg \sin\theta$$ $$ma = F_{applied} - mg \sin\theta$$ $$F_{applied} = ma + mg \sin\theta$$ $$F_{applied} = m(a + g \sin\theta)$$

Using $\sin(37^\circ) \approx 0.6018$:

$$F_{applied} = 24 \, \text{kg} (4.0 \, \text{m/s}^2 + 9.8 \, \text{m/s}^2 \times \sin(37^\circ))$$ $$F_{applied} = 24 (4.0 + 9.8 \times 0.6018)$$ $$F_{applied} = 24 (4.0 + 5.89764)$$ $$F_{applied} = 24 (9.89764)$$ $$F_{applied} = 237.54336 \, \text{N}$$

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