# Consider a situation where a constant force of 25 N acts on an object having a mass of 2 kg for 3...

## Question:

Consider a situation where a constant force of 25 N acts on an object having a mass of 2 kg for 3 seconds. What is the work done by the force?

A) 150 J

B) 50 J

C) 75 J

D) Undetermined - not enough information

E) None of the above

## Work Done:

Work is done on a system if a force is applied on the system and displacement has taken place. Work done is given by the mathematical relation

$$\begin{align} W = F \cdot s \end{align} $$

where {eq}F{/eq} is the force acting on the system and {eq}s{/eq} is the displacement resulting from the applied force.

## Answer and Explanation:

**Data Given**

- Force applied to the object, {eq}(F) = 25 \ \rm N {/eq}

- Mass of the object, {eq}(m) = 2.0 \ \rm kg {/eq}

- Time for which force acts, {eq}(t) = 3.0 \ \rm s {/eq}

The acceleration of the object using Newton's second law

$$\begin{align} F &= ma \\[0.3cm] a &= \frac{F}{m} \\[0.3cm] a &= \frac{25 \ \rm N}{2.0 \ \rm kg} \\[0.3cm] a &= 12.5 \ \rm m/s^2 \end{align} \\ $$

Now, the displacement of the object in the given period is solved using the second equation of kinematics

$$\begin{align} s &= \frac{1}{2} at^2 \\[0.3cm] &= \frac{1}{2} \left( 12.5 \ \frac{\rm m}{\rm s^2}\right) (3 \ \rm s)^2 \\[0.3cm] &= 56.25 \ \rm m \end{align} \\ $$

Finally, the work done by the force is

$$\begin{align} W &= F\cdot s \\[0.3cm] &= 25 \ \rm N ( 56.25 \ \rm m ) \\[0.3cm] &=\boxed{ 1406.25 \ \rm J } \\[0.3cm] \end{align} \\ $$

Thus, the correct option is **E) None of the above**.

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from Physics 101: Help and Review

Chapter 17 / Lesson 4