46. A ball is west of the edge of a ledge and force is applied in the manner in

Question

46. A ball is west of the edge of a ledge and force is applied in the manner in the following Table. The ball weighs 3 kg. There are other forces at work – like gravity and a normal force, which are not in the table (yet?). Assume no friction and no drag.

Force

Magnitude

Direction

X-component

Y-component

Force 1

10 N

0° due East

Force 2

2 N

135° from East

A) What is the resulting acceleration of the ball?

B) After 10 seconds, what would be the velocity of the ball?

C) After ten seconds how much work would be done?

D) Since the ledge potentially interferes with the possible travel of part a. (if the ball is not far enough away and falls off), assume that the velocity when the ball leaves the edge of the cliff is 7.5 m/s.Calculate the magnitude of the Kinetic Energy of the ball at that time.

E) If the ledge is 20 feet high and Forces 1 and 2 are discontinued when the ball moves off the ledge, what is the Potential Energy of the ball leaving the edge?

F) What is the total energy of the ball leaving the edge of the cliff, based on d. and e.?

G) All of the energy is converted to kinetic energy just before the ball hits the ground. What is the velocity of the ball at that time?

H) At ground level there is actually a small puddle into which the ball splashes.Assume inconvenient losses result in 75% efficiency and convert all of the balls remaining kinetic energy into thermal energy.What is the temperature change in the 200 kg of water remaining in the pond?

I) What theory and formulae are necessary for these calculations?

J) How far from the foot of the vertical cliff did the ball hit the ground at the bottom of the cliff?

Force

Magnitude

Direction

X-component

Y-component

Force 1

10 N

0° due East

Force 2

2 N

135° from East

Explanation / Answer

Given,

m = 3 kg ; F1 = 10N ; F2 = 2 N

F1x = F1 cos0 = F1 = 10 N and

F2x = F2 cos(135) = -1.41 N

Fnet = sqrt [ 102 + (-1.41)2 ]= 10.1 N

As the motion of the ball is possible in only horizontal direction, we will consider only horizontal forces resulting in the acceleration.

Fnet = ma

a = Fnet/m = 10.1/3 = 3.37 m/s2

Hence, a = 3.37 m/s2.

(b)given t = 10 s, let v be the velocity. we know that,

v = u + at

v = 0 + 3.37 m/s2 x 10s = 33.7 m/s

Hence, v = 33.7 m/s

c)we know that

Work = F x d

d = v x t = 33.7m/s x 10s = 337 m

work = W = 10.1 N x 337 m = 3403.7 J

Hence, W = 3403.7 J

d)v = 7.5 m/s

we know that, KE = 1/2 m v2 = 0.5 x 3 x 7.5 x 7.5 = 84.38 J

Hence, KE = 84.38 J

e)h = 20 ft = 6.1 m

PE = m g h = 3 x 9.8 x 6.1 = 179.34 J

Hence, PE = 179.34 J

f)total energy would be

E = PE + KE

E = 179.34 + 84.38 = 263.72 J

Hence, E = 263.72 J

g) As per stated,

E = KE

1/2 m v2 = 263.72 J

v = sqrt ( 2 x 263.72 / 3) = 13.26 m/s

Hence, v = 13.26 m/s.

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