A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1. Compute the
work done by the applied force in 10 s,
work done by friction in 10 s,
work done by the net force on the body in 10 s,
change in kinetic energy of the body in 10 s, and interpret your results.

Question

A wire of resistance $ 9 \, \Omega $ is bent to form an equilateral triangle. Then the equivalent resistance across any two vertices will be:

Answer 1

Given Data

Parameter	Value
Mass ($m$)	2 kg
Applied force ($F_\text{app}$)	7 N
Kinetic friction ($\mu_k$)	0.1
Time	10 s
Initial velocity	0 m/s

Step 1: Forces & Acceleration

$$f_k = \mu_k mg = 0.1 \times 2 \times 9.8 = 1.96 \, \text{N}$$ $$F_\text{net} = F_\text{app} - f_k = 7 - 1.96 = 5.04 \, \text{N}$$ $$a = \frac{F_\text{net}}{m} = \frac{5.04}{2} = 2.52 \, \text{m/s}^2$$

Step 2: Kinematics (10 s)

$$v = u + at = 0 + 2.52 \times 10 = 25.2 \, \text{m/s}$$ $$s = ut + \frac{1}{2}at^2 = \frac{1}{2} \times 2.52 \times 10^2 = 126 \, \text{m}$$

Work Calculations

**(a) Applied force work:** $W_\text{app} = F_\text{app} \times s = 7 \times 126 = 882 \, \text{J}$

**(b) Friction work:** $W_f = -f_k \times s = -1.96 \times 126 = -247 \, \text{J}$

**(c) Net force work:** $W_\text{net} = F_\text{net} \times s = 5.04 \times 126 = 635 \, \text{J}$

**(d) ΔKE:** $\Delta K = \frac{1}{2}mv^2 = \frac{1}{2} \times 2 \times (25.2)^2 = 635 \, \text{J}$

Results Summary

Quantity	Value
Applied force work	882 J
Friction work	-247 J
Net force work	635 J
ΔKE	635 J

Interpretation (Work-Energy Theorem)

$W_\text{net} = \Delta KE$: 635 J = 635 J ✓
Total work: $W_\text{app} + W_f = 882 - 247 = 635$ J = $\Delta KE$ ✓
Applied force does more positive work than friction's negative work
Net positive work → kinetic energy gain
Friction dissipates 247 J as heat

Solution and Explanation