A physics student pulls a block of mass m = 16 kg up an incline at a slow consta

Question

A physics student pulls a block of mass m = 16 kg up an incline at a slow constant velocity for a distance of d = 3 m. The incline makes an angle Delta = 26 degree with the horizontal. The coefficient of kinetic friction between the block and the inclined plane is mu_k = 0.2. What is the work W_m done by the student? W_m = At the top of the incline, the string by which she was pulling the block breaks. The block, which was at rest, slides down a distance d = 3 m before it reaches a frictionless horizontal surface. A spring is mounted horizontally on the frictionless surface with one end attached to a wall The block hits the spring, compresses it a distance L = 0.6 m, then rebounds back from the spring, retraces its path along the horizontal surface, and climbs up the incline. What is the speed v of the block when it first reaches the horizontal surface? v = What is the spring constant k of the spring? k = How far up the incline d_1 does the block rebound? d_1 =

Explanation / Answer

1) the student does work against gravity and against friction

Work done by student against gravity = mgh =16*9.8*3sin26=206.21 Joules

Work done by student against friction = (mu k)*mgcos26*3 = 0.2*16*9.8*cos26*3 =84.56 Joules

Therefore, total work done by student=206.21+84.56=290.77 Joules

2)initial potential energy of block when it is at top of incline = mgh=16*9.8*3sin26=206.21 Joule

Energy lost due to friction,while sliding down the incline=84.56Joules

Therefore, energy of mass when it reaches horizontal surface=206.21-84.56 =121.65Joules

This is the total energy of block at horizontal surface,which is also equal to the kinetic energy of block

Velocity of block at this time =sqrt(2*121.65/16)=3.9 m/s

3) compression of spring = 0.6m

Energy stored on spring=energy transferred to spring by block =kinetic energy of block

Let spring constant be k

(1/2)*k*0.6² = 121.65

Therefore, k = 675.83 N/m

4)let the rebound distance be d1

Kinetic energy of block just before it starts climbing up the incline=121.65 J

Energy consumed by friction = 0.2*16*9.8*cos26*d1 =28.19d1

Potential Energy gained in reaching the height = mgh=16*9.8*d1*sin26 =68.74d1

Therefore, 121.65 = 28.19d1 + 68.74d1

Therefore, d1=1.26 m

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