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State Newton's second law of motion. Two particles of mass 15 kg and 20 kg are connected by a taunt light inextensible string which passes over a smooth light pulley. The system shown in fig below is released from rest. Find: i. The common acceleration of the particles ii. The tension of the string

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5 parts
b(i)

1.4 m/s2\text{}1.4 \text{ m/s}^2\text{}

b(ii)

168 N\text{}168 \text{ N}\text{}

c(i)

55.15∘\text{}55.15^\circ\text{}

c(ii)

167.28 N\text{}167.28 \text{ N}\text{}

c(iii)

1.0 m/s2 opposite to the direction of motion\text{}1.0 \text{ m/s}^2 \text{ opposite to the direction of motion}\text{}

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Part a

State Newton's second law of motion

Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. It is commonly expressed as Fnet=maF_{net} = ma, where FnetF_{net} is the net force, mm is the mass, and aa is the acceleration.

Part b(i)

1.4 m/s2\text{}1.4 \text{ m/s}^2\text{}

The common acceleration of the particles

Two particles of mass 15 kg and 20 kg are connected by a string over a smooth light pulley.

  1. Identify the forces acting on each mass.

    Let m1=15 kgm_1 = 15 \text{ kg} and m2=20 kgm_2 = 20 \text{ kg}. Let TT be the tension in the string and g=9.8 m/s2g = 9.8 \text{ m/s}^2 be the acceleration due to gravity. Since m2>m1m_2 > m_1, m2m_2 will accelerate downwards and m1m_1 will accelerate upwards.

    For m1m_1 (15 kg, moving up):

    T−m1g=m1a(1)T - m_1 g = m_1 a \quad (1)

    For m2m_2 (20 kg, moving down):

    m2g−T=m2a(2)m_2 g - T = m_2 a \quad (2)

  2. Add the two equations to eliminate TT and solve for aa.

    (T−m1g)+(m2g−T)=m1a+m2a(T - m_1 g) + (m_2 g - T) = m_1 a + m_2 a m2g−m1g=(m1+m2)am_2 g - m_1 g = (m_1 + m_2) a a=(m2−m1)gm1+m2a = \frac{(m_2 - m_1) g}{m_1 + m_2}

  3. Substitute the given values into the equation for aa.

    a=(20 kg−15 kg)×9.8 m/s215 kg+20 kga = \frac{(20 \text{ kg} - 15 \text{ kg}) \times 9.8 \text{ m/s}^2}{15 \text{ kg} + 20 \text{ kg}} a=(5 kg)×9.8 m/s235 kga = \frac{(5 \text{ kg}) \times 9.8 \text{ m/s}^2}{35 \text{ kg}} a=49 N35 kga = \frac{49 \text{ N}}{35 \text{ kg}} a=1.4 m/s2a = 1.4 \text{ m/s}^2

    The common acceleration of the particles is 1.4 m/s2\boxed{\text{}1.4 \text{ m/s}^2\text{}}.

Part b(ii)

168 N\text{}168 \text{ N}\text{}

The tension of the string

  1. Substitute the calculated acceleration aa into equation (1).

    T−m1g=m1aT - m_1 g = m_1 a T=m1g+m1aT = m_1 g + m_1 a T=m1(g+a)T = m_1 (g + a)

  2. Substitute the values for m1m_1, gg, and aa.

    T=15 kg(9.8 m/s2+1.4 m/s2)T = 15 \text{ kg} (9.8 \text{ m/s}^2 + 1.4 \text{ m/s}^2) T=15 kg(11.2 m/s2)T = 15 \text{ kg} (11.2 \text{ m/s}^2) T=168 NT = 168 \text{ N}

    The tension in the string is 168 N\boxed{\text{}168 \text{ N}\text{}}.

Part c(i)

55.15∘\text{}55.15^\circ\text{}

What angle does the strap make with the horizontal?

A woman pulling her 20 kg suitcase at a constant speed.

Given: Mass m=20 kgm = 20 \text{ kg}, pulling force FP=35 NF_P = 35 \text{ N}, frictional force Ff=20 NF_f = 20 \text{ N}. The suitcase moves at a constant speed, so the net acceleration is 0.

  1. Analyze the horizontal forces.

    Since the suitcase moves at a constant speed, the net horizontal force is zero. The horizontal component of the pulling force balances the frictional force.

    FPcos⁡θ−Ff=0F_P \cos \theta - F_f = 0 FPcos⁡θ=FfF_P \cos \theta = F_f

  2. Solve for θ\theta.

    cos⁡θ=FfFP\cos \theta = \frac{F_f}{F_P} cos⁡θ=20 N35 N\cos \theta = \frac{20 \text{ N}}{35 \text{ N}} cos⁡θ=47\cos \theta = \frac{4}{7} θ=arccos⁡(47)\theta = \arccos\left(\frac{4}{7}\right) θ≈55.15∘\theta \approx 55.15^\circ

    The angle the strap makes with the horizontal is 55.15∘\boxed{\text{}55.15^\circ\text{}}.

Part c(ii)

167.28 N\text{}167.28 \text{ N}\text{}

What normal force does the ground exert on the suitcase?

  1. Analyze the vertical forces.

    Since there is no vertical acceleration, the net vertical force is zero. The normal force (FNF_N) and the vertical component of the pulling force (FPsin⁡θF_P \sin \theta) balance the gravitational force (mgmg).

    FN+FPsin⁡θ−mg=0F_N + F_P \sin \theta - mg = 0 FN=mg−FPsin⁡θF_N = mg - F_P \sin \theta

  2. Substitute the values for mm, gg, FPF_P, and θ\theta.

    FN=(20 kg×9.8 m/s2)−(35 N×sin⁡(55.15∘))F_N = (20 \text{ kg} \times 9.8 \text{ m/s}^2) - (35 \text{ N} \times \sin(55.15^\circ)) FN=196 N−(35 N×0.8206)F_N = 196 \text{ N} - (35 \text{ N} \times 0.8206) FN=196 N−28.72 NF_N = 196 \text{ N} - 28.72 \text{ N} FN=167.28 NF_N = 167.28 \text{ N}

    The normal force exerted by the ground is 167.28 N\boxed{\text{}167.28 \text{ N}\text{}}.

Part c(iii)

1.0 m/s2 opposite to the direction of motion\text{}1.0 \text{ m/s}^2 \text{ opposite to the direction of motion}\text{}

If the strap breaks, what would be the magnitude and direction of the acceleration of the suitcase?

  1. Identify the forces acting on the suitcase after the strap breaks.

    When the strap breaks, the pulling force FPF_P becomes zero. The suitcase is still moving, so the frictional force Ff=20 NF_f = 20 \text{ N} will act opposite to the direction of motion.

  2. Apply Newton's second law in the horizontal direction.

    The net force is the frictional force, acting opposite to the direction of motion.

    Fnet=maF_{net} = ma −Ff=ma-F_f = ma a=−Ffma = -\frac{F_f}{m}

  3. Calculate the acceleration.

    a=−20 N20 kga = -\frac{20 \text{ N}}{20 \text{ kg}} a=−1.0 m/s2a = -1.0 \text{ m/s}^2

    The magnitude of the acceleration is 1.0 m/s21.0 \text{ m/s}^2. The negative sign indicates that the acceleration is in the direction opposite to the initial motion.
    The magnitude and direction of the acceleration are 1.0 m/s2 opposite to the direction of motion\boxed{\text{}1.0 \text{ m/s}^2 \text{ opposite to the direction of motion}\text{}}.

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