What force is required to give a mass of 40 kg (a) acceleration of \( 3 \mathrm{~m} / \mathrm{s}^{2} \) (b) speed of \( 48 \mathrm{~km} / \mathrm{h} \) within 1 sec (c) speed of \( 75 \mathrm{~m} / \mathrm{s} \) within 5 sec (d) distance of 55 m within 8 sec

To find the force required for each scenario, we can use Newton's second law of motion, which states that \( F = m \cdot a \), where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. ### (a) Acceleration of \( 3 \, \mathrm{m/s^2} \) Given: - Mass \( m = 40 \, \mathrm{kg} \) - Acceleration \( a = 3 \, \mathrm{m/s^2} \) Using the formula: \[ F = m \cdot a = 40 \, \mathrm{kg} \cdot 3 \, \mathrm{m/s^2} = 120 \, \mathrm{N} \] ### (b) Speed of \( 48 \, \mathrm{km/h} \) within 1 sec First, we need to convert the speed from \( \mathrm{km/h} \) to \( \mathrm{m/s} \): \[ 48 \, \mathrm{km/h} = \frac{48 \times 1000 \, \mathrm{m}}{3600 \, \mathrm{s}} = \frac{48000}{3600} \approx 13.33 \, \mathrm{m/s} \] Now, we can find the acceleration required to reach this speed in 1 second: \[ a = \frac{\Delta v}{\Delta t} = \frac{13.33 \, \mathrm{m/s}}{1 \, \mathrm{s}} = 13.33 \, \mathrm{m/s^2} \] Now, using \( F = m \cdot a \): \[ F = 40 \, \mathrm{kg} \cdot 13.33 \, \mathrm{m/s^2} \approx 533.2 \, \mathrm{N} \] ### (c) Speed of \( 75 \, \mathrm{m/s} \) within 5 sec Using the same approach, we find the acceleration: \[ a = \frac{\Delta v}{\Delta t} = \frac{75 \, \mathrm{m/s}}{5 \, \mathrm{s}} = 15 \, \mathrm{m/s^2} \] Now, using \( F = m \cdot a \): \[ F = 40 \, \mathrm{kg} \cdot 15 \, \mathrm{m/s^2} = 600 \, \mathrm{N} \] ### (d) Distance of 55 m within 8 sec To find the required force, we first need to determine the average speed needed to cover 55 m in 8 seconds: \[ \text{Average speed} = \frac{\text{Distance}}{\text{Time}} = \frac{55 \, \mathrm{m}}{8 \, \mathrm{s}} = 6.875 \, \mathrm{m/s} \] Next, we can find the acceleration needed to reach this average speed from rest. The average speed can also be expressed as: \[ \text{Average speed} = \frac{u + v}{2} \] where \( u = 0 \) (initial speed) and \( v \) is the final speed. Thus: \[ 6.875 = \frac{0 + v}{2} \implies v = 2 \times 6.875 = 13.75 \, \mathrm{m/s} \] Now, we can find the acceleration: \[ a = \frac{\Delta v}{\Delta t} = \frac{13.75 \, \mathrm{m/s}}{8 \, \mathrm{s}} \approx 1.71875 \, \mathrm{m/s^2} \] Finally, using \( F = m \cdot a \): \[ F = 40 \, \mathrm{kg} \cdot 1.71875 \, \mathrm{m/s^2} \approx 68.75 \, \mathrm{N} \] ### Summary of Forces Required: (a) \( 120 \, \mathrm{N} \) (b) \( 533.2 \, \mathrm{N} \) (c) \( 600 \, \mathrm{N} \) (d) \( 68.75 \, \mathrm{N} \)