Question

An 0.80-kg block is held in place against the spring by a 67-N horizontal external force. The external force is removed, and the block is projected with a velocity of 1.2 m/s upon separation from the spring. The block descends a ramp and has a velocity of 1.9 m/s at the bottom. The track is frictionless between points A and B. The block enters a rough section at B, extending to E. The coefficient of kinetic friction over this section is 0.39. The velocity of the block is 1.4 m/s at C. The block moves on to D, where it stops. What is the spring constant of the spring?

Answer 1

Initial energy:
`( U_i = (1)/(2)kx^2 ), ( K_i = 0 ), ( W_{\text{friction, i}} = 0 ).`

At point C:
`( U_C = (1)/(2)kx^2 ), ( K_C = (1)/(2)mv_C^2 ), ( W_{\text{friction, C}} = \mu mgh ).`

At point D:
`( U_D = 0 ), ( K_D = (1)/(2)mv_D^2 ), ( W_{\text{friction, D}} = \mu mgh ).`

To determine the spring constant, we can use the conservation of mechanical energy.

The potential energy stored in the spring when compressed is converted into kinetic energy and work done against friction as the block moves along the track.

The potential energy stored in the spring is given by
`( U = (1)/(2)kx^2 ),`

where ( k ) is the spring constant and ( x ) is the displacement from the equilibrium position.

The kinetic energy of the block is given by
`( K = (1)/(2)mv^2 )`, where ( m ) is the mass and ( v ) is the velocity.

The work done against friction is
`( W_{\text{friction}} = \mu mgh ),` where ( \mu ) is the coefficient of kinetic friction, ( g ) is the acceleration due to gravity, and ( h ) is the height.

The conservation of energy equation is
`( U = K + W_{\text{friction}} )`.Let's denote the compression of the spring as ( x ).

Initially, the block is held against the spring, so
`( U_i = (1)/(2)kx^2 )` and ( K_i = 0 ).

When the block is released, it gains kinetic energy as it separates from the spring, and this kinetic energy is gradually lost due to friction along the incline.

Finally, at point D where the block stops, all the initial potential energy is lost to friction, so ( U_f = 0 ) and
`( K_f = (1)/(2)mv_f^2 ).`