Homework #3 Solutions

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Physics 221 Summer 2012

HOMEWORK #3

Due Monday July 2, 2012

1
James Bond (90 kg), outfitted with perfectly matching skis and skiware, is at the top of a steep slope that a secret spy like him can easily handle. He lets himself go from rest and smoothly slides down the h = 15 m high hill. A big parking lot lies at the bottom of the hill. Since the parking area has been cleared of snow, the friction between the ground and the skis brings our hero to a halt at point D, located at a distance d = 12 m from point C. The descent can be considered frictionless. Take the potential energy to be zero at the bottom of the slope. (a) What is the mechanical energy of James Bond at points A and D? (b) Determine the speed of Bond at position B abd C. (c) What is the work done by friction in the parking lot? (d) Find the magnitude of the average friction force. SOLUTION : (a) At point A EA = mgh = (90 kg) 9.80 m/s2 (15 m) = 13230 J At point B, James Bond is at rest so, EB = 0 (b) Using conservation of energy: EA = EB ⇒ EA = EC ⇒ mgh = mg h 1 + mv 2 ⇒ 2 2 B ⇒ vB = vC = gh = 2gh = 9.80 m/s2 (15 m) = 12.1 m/s 2 9.80 m/s2 (15 m) = 17.1 m/s

1 2 mgh = mvC 2

(c) Using conservation of energy: EC + Wf
C→D

= ED ⇒

Wf

C→D

= ED − EC = -13230 J

where we used the fact that EC = EA . (d) Using the definition of work: Wf = f kd ⇒ fk = Wf C→D

C→D

d

=

13230 J = 1100 N 12 m

1

Physics 221 Summer 2012

HOMEWORK #3

Due Monday July 2, 2012

2
An object of mass m = 3.00 kg is released from rest at a height of h = 5.00 m on an inclined ramp with angle θ = 10.0◦ with the horizontal. At the foot of the ramp and after a horizontal surface of length d = 20.0 m is the tip of a spring of force constant ik = 4000 N/m (see figure). The object slides down the ramp and into the spring, compressing it a distance x before coming momentarily to rest. Find x and describe what happens to the object after it comes to rest. (a) If all surfaces are frictionless. (b) If the coefficient of kinetic friction µk between the horizontal surface and the object is 0.200. SOLUTION : (a) With frictionless surfaces, Energy is conserved. When the spring is completely compressed, kinetic energy is zero, therefore: Einitial = Efinal ⇒ 1 mgh + 0 + 0 = 0 + kx2 + 0 2 2mgh = k 2 (3.00 kg) 9.80 m/s2 (5.00 m) 4000 N/m = 0.271 m



x=

The objects is then pushed back by the spring making it climb the incline to the original hieght. (b) With friction on the horizontal surface, the final energy will be less than initial energy by the magnitude of work lost to friction. A note is that until the spring is completely compressed by a distance x , the object has travelled on the horizontal surface for a distance of d + x . Using conservation of energy thoerem: Einitial + Wf = Efinal ⇒ ⇒ 1 mgh − µk mg d + x = kx 2 2 2µk mg 2mg (µk d − h) 2 x − x + =0 k k x 2 + (0.00294 m) x − 0.0147 m2 = 0

Solving this quadratic equation for x , we get: x = 0.123 m or ((((( m x = −0.120 ( ((

2

Physics 221 Summer 2012

HOMEWORK #3

Due Monday July 2, 2012

3
A skier starts from rest at height H above the center of a rounder hummock of radius R (see figure). There is negligible friction. Find the maximum value of H for which the skier remains in contact with the snow at the peak of the hummock.

SOLUTION: For the skier not to fly off hummock, then it has to have the speed that can satisfy the condition of circular motion at the top. Using the free-body-diagram at that point as shown in the figure. We use newton’s second equation of motion: Fradial = Fnet, radial N − mg = −m

N Fnet = m mg
2 vtop R

2 vtop (1) R From the last equation, the faster the skier at the top, the smaller N . The maximum allowed speed, leads to N = 0 and therefore:

vtop, max =

gR

(2)

The speed at the top of the hummock is also determined my the initial energy. Using conservation of energy: Einitial = Etop 1 2 0 + mgH = mvtop + mgR 2 Rearrange the last equation to solve...
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