A crate with mass M_crate = 110.0 kg is being pulled up a ramp with a constant s

Question

A crate with mass M_crate = 110.0 kg is being pulled up a ramp with a constant speed of 1.50 m/s by a block with a mass of M_block = 60.0 kg. The crate and block are connected by a massless rope that cannot be stretched. The pulley is frictionless and massless. At the instant shown in the figure below, the block has a height of h = 2.00 m above the ground. The incline makes an angle of ? = 25.0 degrees with respect to the ground. a). Explain why the tension in the rope does no work on the block-crate-earth system. b). Just before the block reaches the ground, how much energy of the system has been converted to thermal energy? c). What is the coefficient of kinetic friction between the crate and the ramp? d). How much further will the crate go up the ramp once the block has reached the ground? Assume the collision of the block with the ground is completely inelastic.

Explanation / Answer

a) friction does no work because friction always opposes the motion, and the motion of crate is in opposite direction of the friction force....also, there is no concept of negative work...hence friction does no work as there is no motion in the direction of friction... All the potential energy change will be converted into thermal energy, as the velcity is constant, which implies there is no net force...also change in kinetic energy is zero...so...all the potential energy is lost to friction....Hence total enegy coverted to thermal energy = mgh = (110 + 60) * 9.81 * 2 = 3.335 kJ c)110 g sin 25 + uk * 110 g cos 25 = 60 g solving this we get...uk = 0.135 d) block will go upward because of the initial velocity that it has... = 1.5 m/s net down the incline force on block = 110 g sin 25 + uk * 110 g cos 25 = 60 g so acceleration = 60 g / 110 = 5.35 m/s^2 Now v^2 - u^2 = 2 a s 0 - 2.25 = 2 * -5.35 * s s = 0.21 m

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