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Question: A body of mass \(2{{kg}}\) initially at rest moves under the action of an applied horizontal force o...

A body of mass 2kg2{{kg}} initially at rest moves under the action of an applied horizontal force of 7N7{{N}} on a table with the coefficient of kinetic friction =0.1 = 0.1. Compute the:
(a) Work done by the applied force in 10s10{{s}}
(b) Work done by friction in 10s10{{s}}
(c) Work done by the net force on the body in 10s10{{s}}
(d) Change in kinetic energy of the body in 10s10{{s}}
And interpret your results.

Explanation

Solution

Work done is defined as the product of the magnitude of the displacement and the component of force in the direction of the displacement. When there is friction present on the surface, the force put to do some work is used in overcoming the friction forces and to displace the body.

Formula used:
W=F×SW = F \times S, and
ΔK=12mv2−12mu2\Delta K = \dfrac{1}{2}m{v^2} - \dfrac{1}{2}m{u^2}

Complete step by step solution:
Mass of the body m=2kg{{m = }}2{{kg}}
Force applied on the body F=7N{{F = }}7{{N}}
Coefficient of kinetic friction =0.1 = 0.1
Since the body was initially at rest, therefore its initial velocity will be 00
The time at which the work is to be calculated t=10s{{t}} = 10{{s}}
We know from Newton’s second law of motion:
F=m×a a=Fm\begin{gathered} {{F = m}} \times {{a}} \\\ {{a = }}\dfrac{{{F}}}{{{m}}} \end{gathered}
Putting the values of F{{F}}and m{{m}}, we get:
⇒a=72=3.5m/s2\Rightarrow {{a = }}\dfrac{7}{2} = 3.5{{m/}}{{{s}}^2}
And we know that kinetic frictional force is:
⇒f=μ×g=0.1×2×9.8=1.96\Rightarrow {{f = }}\mu \times g = 0.1 \times 2 \times 9.8 = 1.96
And the acceleration produced by the frictional force will be
⇒a′=−1962=−0.98m/s2\Rightarrow {{a' = }}\dfrac{{ - 196}}{2} = - 0.98{{m/}}{{{s}}^2}
The frictional forces oppose the motion of the body; therefore, they decelerate the body. Hence the acceleration is taken as negative.
The total acceleration of the body will be:
⇒a+a′=3.\Rightarrow {{a + a'}} = 3. a+a′=3.5+(−0.98)=2.52m/s2{{a + a' }} = 3.5 + ( - 0.98) = 2.52{{m/}}{{{s}}^2}
The equation of motion states that distance travelled:
⇒s=ut+12at2=0+12×2.52×(10)2=126m\Rightarrow {{s = ut + }}\dfrac{1}{2}{{a}}{{{t}}^2} = 0 + \dfrac{1}{2} \times 2.52 \times {(10)^2} = 126{{m}}
(a) Work done by the applied force in 10s10{{s}}
⇒W=F×S=7×126=882J\Rightarrow {{W = F}} \times {{S = }}7 \times 126 = 882{{J}}

(b) Work done by the frictional force in 10s10{{s}}
⇒W′=F×S=1.96×126=247J\Rightarrow {{W' = F}} \times {{S = }}1.96 \times 126 = 247{{J}}

(C) Work done by the net force on the body in 10s10{{s}}
Net force will be the total force exerted on the body, which is the difference between the applied force and frictional force.
⇒7+(−1.96)=5.04N\Rightarrow 7 + ( - 1.96) = 5.04{{N}}
Work done by the net force =5.04×126=635J{{ = }}5.04 \times 126 = 635{{J}}

(d) Change in kinetic energy of the body in 10s10{{s}}
From the first equation of motion,
⇒v=u+at=0+2.52×10=25.2m/s2\Rightarrow {{v = u + at = }}0 + 2.52 \times 10 = 25.2{{m/}}{{{s}}^2}
The change in kinetic energy:
⇒ΔK=12mv2−12mu2=12×2×(v2−u2)\Rightarrow \Delta {{K = }}\dfrac{1}{2}{{m}}{{{v}}^2} - \dfrac{1}{2}{{m}}{{{u}}^2} = \dfrac{1}{2} \times 2 \times ({{{v}}^2}{{ - }}{{{u}}^2})
⇒ΔK=(25.2)2−(0)2=635J\Rightarrow \Delta {{K = (}}25.2{{{)}}^2} - {(0)^2} = 635{{J}}

Note: The force applied is utilized to do the work while the friction forces try to oppose the forces applied. Therefore, care must be taken about the signs of the frictional force and the acceleration as the frictional forces try to decelerate the body.