A force of `F=(0.5x+12)N` acts on a particle. If x is in metre, calculate the work done by the force during the displacement of the particle from x = 0 to x = 4 m

Choose the correct options A force F = (0.5x + 12) N acts on a particle. If x is in metres, calculate the work done by the force during the displcement of the particle from x= 0 to x = 4 m

A variable force F=kx^(2) acts on a particle which is initially at rest. Calculate the work done by the force during the displacement of the particle from x=0 m "to" x= 4m . (Assume the constant k=1Nm^(-2) )

A force F = 0.5x + 10 acts on a particle. Here F is in newton and x is in metre. Calculate the work done by the force during the displacement of the particle from x = 0 to x = 2 metre.

A force F = (10 +0.25x ) is acting on a particle moving along positive X direction . Calculate the work done by this force during a displacement from x = 1 m to x = 4 m .

Force acting on a particle varies with x as shown in figure. Calculate the work done by the force as the particle moves from x = 0 to x = 6.0 m.

A block of mass 5 kg initially at rest at the origin is acted upon by a force along the positive X-direction represented by F=(20+5x) N . Calculate the work done by the force during the displacement of the block from x = 0 to x = 4m .

A block of mass 5 kg initially at rest the origin is acted on by a force along the X - positive direction represented by F = (20+5x) N. Calculate the work done by the force during the displacement of the block from x = 0 to x = 4 m .

A force F=(10+0.5 x) acts on a particle in the x-direction. What would be the work done by this force during a displacement from x=0 to x=2m (F is in newton and x in metre)