Question

Applying a force of 20 N to a 0.4 kg puck as it slides along to frictionless surface from rest to 10 ms-1 in 0.2 s. Calculate the work done by the puck.A. 2000 JB. 200 JC. 20 J

Answer 1

Given,

The force applied on the puck, F=20 N

The mass of the puck, m=0.4 kg

The initial velocity of the puck, u=0 m/s

The final velocity of the puck, v=10 m/s

The time duration, t=0.2 s

The work done is given by the product of the force applied to the object and the distance to which the force was applied.

That is,

$W=F\times d\text{ }\rightarrow\text{ (i)}$

Where d is the distance traveled by the puck.

From the equation of motion,

$d=ut+\frac{1}{2}at^2$

Where a is the accleration of the puck due to the applied force F.

But, the puck was at rest initially. Therefore,

$d=\frac{1}{2}at^2\text{ }\rightarrow\text{ (ii)}$

From Newton's second law of motion,

$\begin{gathered} F=ma \\ \Rightarrow a=\frac{F}{m}\text{ }\rightarrow\text{ (iii)} \end{gathered}$

On substituting the equation (iii) in equation (ii),

$\begin{gathered} d=\frac{1}{2}\times\frac{F}{m}\times t^2 \\ =\frac{Ft^2}{2m}\text{ }\rightarrow(\text{iv)} \end{gathered}$

On substituting equation (iv) in equation (i),

$\begin{gathered} W=F\times\frac{Ft^2}{2m} \\ =\frac{F^2t^2}{2m} \end{gathered}$

On substituting the known values in the above equation,

$\begin{gathered} W=\frac{20^2\times0.2^2}{2\times0.4} \\ =20\text{ J} \end{gathered}$

Therefore, the work done by the puck is 20 J