A67.0-kg runner has a speed of 5.80 m/s at one instant during a long-distance ev

Question

A67.0-kg runner has a speed of 5.80 m/s at one instant during a long-distance event. (a) What is the runner's kinetic energy at this instant? (b) If he doubles his speed to reach the finish line, by what factor does his kinetic energy change? A mechanic pushes a 3.20. 10^3-kg car from rest to a speed of v, doing 5, 400 J of work in the process. During this time, the car moves 28.0 m. Neglecting friction between car and road, find v and the horizontal force exerted on the car. (a) the speed v (b) the horizontal force exerted on the car (Enter the magnitude.) A 74-kg base runner begins his slide into second base when he is moving at a speed of 4.2 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base. (a) How much mechanical energy is lost due to friction acting on the runner? (b) How far does he slide? A 0.24-kg stone is held 1.1 m above the top edge of a water well and then dropped into it. The well has a depth of 4.9 m. (a) Taking y = 0 at the top edge of the well, what is the gravitational potential energy of the stone-Earth system before the stone is released? (b) Taking y = 0 at the top edge of the well, what is the gravitational potential energy of the stone-Earth system when it reaches the bottom of the well?

Explanation / Answer

problem 1:

kinetic energy
KE= 1/2MV^2 = 0.5x67x5.80^2 = 1126.94 J

If he doubles his speed to reach the finish line, by what factor does his kinetic energy change

=  KE new = 4 (KE old) = 4x1126.94 = 4507.76 J

problem 2:

we know, W = F*d
so F = W/d = 5,400 / 28 = 192.857 N

a = F/m = 192.857 / 3200= 0.06 m/s^2 ...........................................ans a)

v^2 = u^2 + 2*a*s = 0 + 2*0.06*28

or, v = 1.83 m/s ................................................................ans b)

problem 3:

The mechanical energy "lost" equals the change in kinetic energy = 1/2*m*v^2 = 1/2*74.0kg*(4.2m/s)^2 = 652.68J

b) Work = F*d where F= mu*m*g so W = mu*m*g*x
So x = W/(mu*m*g) = 652.68J/(0.70*74kg*9.8m/s^2) = 1.286 m

problem 4:

The gravitational potential energy is the energy change that would occur between two points due to gravity and can be written as follows:

PE = m * g * h; PE = potential energy, m = mass =0.24 kg, g = gravity = 9.81m/s^2, h = distance between the two points. For part a, the two points of reference are the stone's current location and the bottom of the well which is considered the zero location in our reference system. This gives:

PE = m * g * h; h = 1.1m + 4.9m = 6.0m. Plugging in the values gives:
PE = 0.24kg * 9.81m/s^2 * 4.9m
= gravitational potential energy of the stone before it is released

For part b when the stone reaches the bottom of the well the stone it is now at the zero point in our system. This means h = 0 giving:

PE = m * g * h; h = 0
PE = 0 = potential gravitational energy of the stone in the bottom of the well

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