A block of mass m moving with a velocity v_0 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to the work-energy theorem,the work done by all forces acting on a body is equal to the change in its kinetic energy.For observer $A$ (g

Vedclass · Accepted Answer

$ - \frac{1}{2}mv_0^2$

Vedclass · Answer

(B) According to the work-energy theorem,the work done by all forces acting on a body is equal to the change in its kinetic energy.For observer $A$ (ground frame),the initial kinetic energy of the block is $K_i = \frac{1}{2}mv_0^2$ and the final kinetic energy is $K_f = 0$ (since the block comes to rest).The only force doing work on the block in the horizontal direction is the spring force $(F_s)$.Therefore,$W_{spring} = K_f - K_i = 0 - \frac{1}{2}mv_0^2 = -\frac{1}{2}mv_0^2$.Thus,the work done by the spring force as observed by $A$ is $-\frac{1}{2}mv_0^2$.

Similar Questions

When a long spring of $4 \ cm$ is stretched by $1 \ cm$,the potential energy stored in the spring is $U$. If it is stretched by $4 \ cm$,the potential energy stored in it is (in $U$)

$A$ spring of force constant $K$ is first stretched by distance $a$ from its natural length and then further by distance $b$. The work done in stretching the part $b$ is .............

In stretching a spring by $2 \, cm$,the energy stored is $U$. If the spring is stretched by an additional $10 \, cm$,the total energy stored will be:

$A$ spring with a force constant of $800 \ N/m$ is stretched by $5 \ cm$. Find the work done in stretching it from $5 \ cm$ to $15 \ cm$ in $J$.

Show that the law of conservation of mechanical energy is obeyed by pulling or compressing the block tied at the end of a spring.

Vedclass Test Series

Exam Paper Generator

Online Exam Module