Find the change in the kinetic energy of the block on the incline as it moves a

Question

Find the change in the kinetic energy of the block on the incline as it moves a distance of 25.4 m up the incline if the system starts from rest. Answer in units of kJ

Two blocks (with masses 35.3 kg and 61 kg) are connected by a string as shown. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between block and machine is 0.224 .The acceleration of gravity is 9.8 m/s2 . Find the change in the kinetic energy of the block on the incline as it moves a distance of 25.4 m up the incline if the system starts from rest. Answer in units of kJ

Explanation / Answer

There are four forces acting on the block on the incline: gravity (its weight mg) acting vertically downward; the tension T in the string, acting parallel to and up the incline toward the pulley; the normal force N, acting perpendicular to and away from the incline; and the frictional force f, acting parallel to and down the incline (assuming that the sliding block is not moving down the incline; but we're told the system starts out at rest, so this statement is justified).

Adopt a coordinate system in which the +x axis points parallel to and up the incline (so T is in the +x direction and f is in the -x direction), and +y points in the same direction as N. Let ? denote the angle that the incline makes with the horizontal.

Resolve the gravity force into components parallel to and perpendicular to the x-axis. These have magnitude mg sin(?) and mg cos(?), respectively.

The sliding block has no net acceleration perpendicular to the incline, so the net force in this direction is zero:

N - mg cos(?) = 0
N = mg cos(?)

As usual, we will use the empirical friction law f = ?N = ?mg cos(?)

The equation for the net force on the sliding block parallel to the incline is

F = ma = T - mg sin(?) - f

Now consider the hanging mass. It is acted upon by two forces: gravity (ITS weight) and T. The string does not stretch, so T has the same magnitude for both blocks; the pulley simply changes its direction. And if the string doesn't stretch (or slacken), the accelerations of the two masses (as well as speeds at any given moment) are the same. This imposes a sense of direction: since we took "up the incline" to be the positive direction, we must take the direction of "a" for the hanging mass to be downward. So for the hanging block we have

Ma = Mg - T

where M is the mass of the (heavier) hanging block.

Solving this for T, we get T = Mg - Ma.

Now substitute for T and f in the force equation for the sliding block:

ma = Mg - Ma - mg sin(?) - ?mg cos(?)

Solve for a.

The force pulling the sliding block is constant in time, so the work it does in moving the block through a distance d is (ma)(d). You are given a value for d. This work is equal to the change in the kinetic energy of the sliding block

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