question is attached. A 5.40-kg block is set into motion up an inclined plane wi
ID: 2113237 • Letter: Q
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question is attached.
A 5.40-kg block is set into motion up an inclined plane with an initial speed of vi = 7.40 m/s (see figure below). The block comes to rest after traveling d = 3.00 m along the plane, which is inclined at an angle of theta = 30.0 degree to the horizontal. For this motion, determine the change in the block's kinetic energy. J For this motion, determine the change in potential energy of the block-Earth system. J Determine the friction force exerted on the block (assumed to be constant). N What is the coefficient of kinetic friction?Explanation / Answer
the initial kinetic energy is
Ekin = 1/2 mv^2 = 1/2 *5*7,6^2 = 144,4 Nm (potential energy = 0 here)
the potential energy = mgh = 5*9,81*3*sin30 = 73,57 Nm (3*sin30 = height where block comes to rest, kinetic energy = 0 here)
The difference is the friction work or energy:
70,83 Nm = u*m*g*cos30*s = u*5*9,81*cos30*3
u = 70,83/(5*9,81*cos30*3)
u = 0,555 = coefficient of kinetic friction
frictional force = u*m*g*cos30 = 23,57 N
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