Question

The velocity of a body of mass 20 kg decreases from \[20{\text{ m}}{{\text{s}}^{ - 1}}\] to \[5{\text{ m}}{{\text{s}}^{ - 1}}\] in a distance of 100 m. Find the Force on the body.
A. -27.5 N
B. -47.5 N
C. -37.5 N
D. -67.5 N

Answer 1

Hint: In this question, we have to find the value of the force acting on the body. For this, we need to use the equation of motion and then the formula of force to get the desired result. Newton’s second law plays a significant role in solving this example.

Formula used:
The formula of equation of a motion is,
\[{v^2} = {u^2} + 2as\]
Where, \[v\] is the final velocity, \[u\] is the initial velocity, \[a\] is the acceleration and \[s\] is the displacement.
Also, according to newton’s second law, we can say that,
\[F = m \times a\]
Where, \[F\] is the force, \[m\] is the mass of an object and \[a\] is the acceleration.

Complete step by step solution:
Let us find the value of acceleration using the equation of motion. The equation of a motion is given by
\[{v^2} = {u^2} + 2as\]
But \[v = 5{\text{ m/s, u}} = {\text{20 m/s, s}} = 100{\text{ }}m\]
Thus, we get
\[{\left( 5 \right)^2} = {\left( {20} \right)^2} + 2a\left( {100} \right)\]
\[\Rightarrow 25 = 400 + 200a\]

By simplifying, we get
\[25 - 400 = 200a\]
\[ \Rightarrow - 375 = 200a\]
\[\Rightarrow \dfrac{{ - 375}}{{200}} = a\]
\[\Rightarrow a = - 1.875{\text{ m/}}{{\text{s}}^2}\]

Now, we will find the value of force acting on a body. According to Newton's second law, we can say that the force is the product of the mass and acceleration of a body.
\[F = m \times a\]
\[\Rightarrow F = 20 \times \left( { - 1.875} \right)\]
By simplifying, we get
\[F = - 37.5{\text{ N}}\]
Hence, the force acting on a body is -37.5 N.

Therefore, the AC motors have slip rings (C).

Note: Many students may get confused with Newton's laws of motion. We will get the wrong result if we apply Newton's law here. Also, it is necessary to calculate the accurate value of acceleration from the equation of motion. Then only we will get the desired result.