# An object whose mass is 0.092 kg is initially at rest and then attains a speed of 75.0 m/s in...

## Question:

An object whose mass is 0.092 kg is initially at rest and then attains a speed of 75.0 m/s in 0.028 s. What average net force acted on the object during this time interval?

A. {eq}\rm 4.9 \times 10^2\ N {/eq}.

B. {eq}\rm 1.2 \times 10^2\ N {/eq}.

C. {eq}\rm 2.5 \times 10^2\ N {/eq}.

D. {eq}\rm 2.8 \times 10^2\ N {/eq}.

E. Zero.

## Classical Mechanics:

Contemplating apparent-sized objects and comparatively larger objects, the theoretical explanation, and proofs regarding the motion of these objects in a detailed manner are known as classical mechanics.

## Answer and Explanation: 1

Contemplating the kinematics equation, the final velocity {eq}v {/eq} of an object is expressed in terms of initial velocity {eq}u {/eq}, time {eq}t {/eq}, and acceleration {eq}a {/eq} of the object. Consequently, the expression is given by,

{eq}v=u+at\text{ }\ \ \ \ \quad \left( 1 \right) {/eq}

The object is initially at rest. As a result, the initial velocity {eq}u {/eq} of the object will be {eq}0\text{ }{\text{m}}/{\text{s}}\; {/eq}.

Further, rearrange equation 1 and substitute the values of initial velocity {eq}u=0\text{ }{\text{m}}/{\text{s}}\; {/eq}, time {eq}t=0.028\text{ s} {/eq}, and final velocity {eq}u=75.0\text{ }{\text{m}}/{\text{s}}\; {/eq} to obtain the acceleration {eq}a {/eq} of the object.

{eq}\begin{aligned} a&=\dfrac{v-u}{t} \\ &=\dfrac{75.0\text{ }{\text{m}}/{\text{s}}\;-0\text{ }{\text{m}}/{\text{s}}\;}{0.028\text{ s}} \\ &=\dfrac{75.0\text{ }{\text{m}}/{\text{s}}\;}{0.028\text{ s}} \\ &\approx 2678.5\text{ }{\text{m}}/{{{\text{s}}^{2}}}\; \\ \end{aligned} {/eq}

Now, in order to determine the net force {eq}F {/eq} acting on the object, multiply the mass {eq}m {/eq} of the object and the acceleration {eq}a {/eq} of the object. Substitute the values of mass {eq}m=0.092\text{ kg} {/eq}, and acceleration {eq}a=2678.5\text{ }{\text{m}}/{{{\text{s}}^{2}}}\; {/eq} in the expression to obtain the net force {eq}F {/eq}.

{eq}\begin{aligned} F&=ma \\ &=\left( 0.092\text{ kg} \right)\left( 2678.5\text{ }{\text{m}}/{{{\text{s}}^{2}}}\; \right) \\ &\approx 246.422\text{ N} \\ &\approx \text{2}\text{.5}\times \text{1}{{\text{0}}^{2}}\text{ N} \\ \end{aligned} {/eq}

**Therefore, option C is valid. **

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Chapter 7 / Lesson 10Learn to define physics as a subject. Discover the many branches of physics. Find out the fundamentals and other important topics in the branches of physics.