Question

8.2

A block of mass m = 5.10 kg is released from rest from point and slides on the frictionless track shown in the figure below. (Let h_a = 7.40 m.)

(a) Determine the block's speed at points and .

vB	=  m/s
vC	=  m/s

(b) Determine the net work done by the gravitational force on the block as it moves from point to point .
J

Answer 1

mass of block (m) = 5.10kg. height at point A (ha = 7.40m. height at point B (hb) = 3.20m. height at point c (hc) = 2m. a), We have to find the block's speed at points B and c. Use work energy theonem , Wg = ARE Wosk done by gravity = change in kinetic energy first, block's speed at point B, Wg = mg (ha-ho) Wg = 1/2 mu So, I m = mg (ha-ho) [:g = 9.8 m 1sec?] 4² = 2mg (ha-ho) m 2 x 500 x 9.88 (7.40-3. 20) 5 to 12 = 82.32 V = 582.32 = 9.07 misec.) Now block's speed at point c, Wg = / m² = mg (ha-he)

2 mg (ha-ha) = 2 x 9.8% (7.40-2) 2 = 105.84 & = 105.84 V = 10.28 msec) by Net workdone by gravitational force on the block as it moves from point A to point c, Wg = mg (ha-he) Wg = 5.1089.8X (7.40-2) Wg = 5.1089.885.40 Wg = 269.892 J