Question

: A force of 30 N acts on a body of mass 2.0 kg starting from rest up to a distance of 3.0 m. Then, the force reduces to 15 N and acts in the same direction up to 2.0 m. Calculate the final kinetic energy of the body.

Answer 1

Hint: We can use the formula for Newton’s second law and kinematical equation and relating final velocity, initial velocity, acceleration and displacement.
Formula used:
\[F = ma\]
Here, \[F\] is the applied force, \[m\] is the mass of the body and \[a\] is the acceleration.
\[v^2 = u^2 + 2as\]
Here, \[v\] is the final velocity, \[u\] is the initial velocity, \[a\] is the acceleration and \[s\] is the distance covered.
\[{\text{kinetic energy}} = \dfrac{1}{2}mv^2 \]

Complete step by step answer:
Calculate the acceleration of the body up to the distance 3.0 m, keeping in mind that the body has started from the rest.
We have given the applied force and mass of the body. Therefore, we can use Newton’s second law to determine the acceleration produced in the body.
Newton’s second law states that the force exerted on the body is equal to the product of mass of the body and acceleration of the body.
\[F = ma\]
Here, \[m\] is the mass of the body and \[a\] is the acceleration in the body.
To determine the acceleration of the body, rearrange the above equation for \[a\] as below,
\[a = \dfrac{F}{m}\]
Substitute 30 N for \[F\] and 2.0 kg for \[m\] in the above equation.
\[a = \dfrac{{30\,N}}{{2.0\,kg}}\]
\[\therefore a = 15\,ms^{ - 2} \]
Now, use the kinematical equation relating final velocity, initial velocity, acceleration and displacement as follows,
\[v^2 = u^2 + 2as\]
Here, \[v\] is the final velocity, \[u\] is the initial velocity, \[a\] is the acceleration and \[s\] is the distance covered.
Since the body is started from the rest, substitute \[u = 0\,ms^{ - 1} \] in the above equation.
\[v^2 = \left( {0\,ms^{ - 1} } \right)^2 + 2as\]
\[v^2 = 2as\]
\[\Rightarrow v = \sqrt {2as} \]
Substitute \[15\,ms^{ - 2} \] for \[a\] and 3.0 m for \[s\] in the above equation.
\[v = \sqrt {2\left( {15\,ms^{ - 2} } \right)\left( {3.0\,m} \right)} \]
\[v = \sqrt {90\,m^2 s^{ - 2} } \]
\[v = 9.5\,ms^{ - 1} \]
This is the final velocity at the end of 3.0 m produced by 30 N force. This final velocity of the body produced by 30 N force is the initial velocity of the body produced by the 15 N force.
Determine the acceleration of the body produced by 15 N force as follows,
\[a = \dfrac{F}{m}\]
Substitute 15 N for \[F\] and 2.0 kg for \[m\] in the above equation.
\[a = \dfrac{{\,15\,N}}{{2.0\,kg}}\]
\[\therefore a = 7.5\,ms^{ - 2} \]
We can calculate the final velocity of the body produced by the action of 15 N force at the end of 2 m distance using the kinematical equation below.
\[v^2 = u^2 + 2as\]
Substitute \[9.5\,ms^{ - 1} \] for \[u\], \[7.5\,ms^{ - 2} \]for \[a\] and 2.0 m for \[s\]in the above equation.
\[v^2 = \left( {9.5\,ms^{ - 1} } \right)^2 + 2\left( {7.5\,ms^{ - 2} } \right)\left( {2.0\,m} \right)\]
\[v^2 = 90.25\,m^2 s^{ - 2} + 30\,m^2 s^{ - 2} \]
\[v^2 = 120.25\,m^2 s^{ - 2} \]
\[\Rightarrow v = 10.96\,ms^{ - 1} \]
The final kinetic energy of the body at the end of 3.0 m distance is,
\[K_f = \dfrac{1}{2}mv^2 \]
Substitute 2.0 kg for m and \[10.96\,ms^{ - 1} \]
 for v in the above equation.
\[K_f = \dfrac{1}{2}\left( {2.0\,kg} \right)\left( {10.96\,ms^{ - 1} } \right)^2 \]
\[K_f = \dfrac{1}{2}\left( {2.0\,kg} \right)\left( {10.96\,ms^{ - 1} } \right)^2 \]
\[\therefore K_f = 120\,J\]
Therefore, the final kinetic energy of the body is 120 J.

Note:Within the path of the body, the force is not constant. One may incorrectly take the force as constant and will use a kinematic equation to determine the final velocity of the body at the end of total 5.0 m distance.