Question
A body of mass 2kg initially at rest moves under the action of an applied horizontal force of 7N on a table with coefficient of kinetic friction = 0.1. Compute the: Change in kinetic energy of the body in 10s.
✓
Answer
M = 2kg Applied force F = 7N Coefficient of kinetic friction $\mu = 0.1$ Normal reaction is N = mg = 2 × 9.8 = 19.6N Hence, force or friction is $\text{f} = \mu\text{N} = 1.96\text{N}$ 
Total force = F - f = 7 - 1.96 = 5.04N
Acceleration of body is, $\text{a}=\Big(\frac{\text{F}-\text{f}}{\text{m}}\Big)$ $=\frac{5.04}{2}\simeq2.5\text{ms}^{-2}$ Displacement of body in time t is, $\text{x}=\frac{1}{2}\text{at}^2$ Int = 10s $\text{x}=\frac{1}{2}\times2.5\times10^2$ $=125\text{m}$ Change in kinetic energy = Net work done = 630J
Total force = F - f = 7 - 1.96 = 5.04NAcceleration of body is, $\text{a}=\Big(\frac{\text{F}-\text{f}}{\text{m}}\Big)$ $=\frac{5.04}{2}\simeq2.5\text{ms}^{-2}$ Displacement of body in time t is, $\text{x}=\frac{1}{2}\text{at}^2$ Int = 10s $\text{x}=\frac{1}{2}\times2.5\times10^2$ $=125\text{m}$ Change in kinetic energy = Net work done = 630J
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