Question

1. A 67-kg base runner begins his slide into second base when he is moving at a speed of 4.5 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?

2. A 0.32-kg stone is held 1.4 m above the top edge of a water well and then dropped into it. The well has a depth of 5.1 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?

(c) What is the change in gravitational potential energy of the system from release to reaching the bottom of the well?

3. In a control system, an accelerometer consists of a 4.91-g object sliding on a calibrated horizontal rail. A low-mass spring attaches the object to a flange at one end of the rail. Grease on the rail makes static friction negligible, but rapidly damps out vibrations of the sliding object. When subject to a steady acceleration of 0.824g, the object should be at a location 0.515 cm away from its equilibrium position. Find the force constant of the spring required for the calibration to be correct.

4. A 48-kg pole vaulter running at 12 m/s vaults over the bar. Her speed when she is above the bar is 1.7 m/s. Neglect air resistance, as well as any energy absorbed by the pole, and determine her altitude as she crosses the bar.

Answer 1

a) The mechanical energy is lost due to friction acting on the runneir Apply work-energy theorem to cal culate the energy lose Here, mis the mass of runner and v is the final speed, and vois the initi al speed W m(0)(05 (67kg) (4.5m/s)-678.375J(o-6.9x102J (b)The di stance he slides is, Therefore, the distance travelled by runner is, -678.375J fn g (0.7)(67kg)(08m/s F 1.47m(or)1.5m