A 5-kg block is pushed down an inelined ramp with an initial velocity, vo. The b

Question

A 5-kg block is pushed down an inelined ramp with an initial velocity, vo. The block slid down the ramp k 500 N/m. The ramp is inclined at an angle of 30° above the horizontal. On hitting the spring, the block compresses the spring through a maximum distance of 35 , through a distance d 0.4 m, before colliding with a spring of force con cm before momentarily coming to rest. The coefficient of kinetic friction between the block and the slope is 0.25 Use energy methods to calculate this initial velocity vo with which the block was pushed 30° 4. A block of mass mi 12 kg on a rough 30°-inclined plane is connected to a 2-kg mass (m by a string of negligible mass that passes over a pulley that is shaped like a ring. This 2-kg pulley has a radius of 20 cm. The string does not slip on the pulley and causes the pulley t rotate about a fixed horizontal axle through its center of mass. When this system is released from rest, the blocks move at a uniform linear acceleration. The coefficient of kinetic friction between block m, and the inclined plane is = 0.15. Assume that the blocks do not hit the pulley. (a) Find the magnitude and direction of the linear acceleration of the kg block. (b) Calculate the tension Ti, in the string supporting m, and tension Ts, in the string that supports m2. (c) What is the total angular momentum of the system when the hanging block moves from rest through a vertical distance of 0.20 m? (d) Wha pulley? t is the net torque on the system that results in the motion of the blocks around the

Explanation / Answer

3] Work done by all forces = change in KE

mg(d+x) sin 30 degree - umg(d+x) cos 30 degree - 0.5kx^2 = 0 - 0.5mv0^2

5*9.8*(0.4+0.35)*sin 30 degree - 0.25*5*9.8*(0.4+0.35)*cos 30 degree - 0.5*500*0.35^2 = 0.5*5v0^2

  -20.2 = - 0.25 v0^2

v0 = sqrt(20.2/0.25)

= 8.99 m/s answer

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