1-) Suppose the ski patrol lowers a rescue sled and victim, having a total mass
ID: 249264 • Letter: 1
Question
1-) Suppose the ski patrol lowers a rescue sled and victim, having a total mass of 95.0 kg, down a = 66.0° slope at constant speed, as shown in Figure 6.22. The coefficient of friction between the sled and the snow is 0.100.
(a) How much work is done by friction as the sled moves 30.0 m along the hill?
J
(b) How much work is done by the rope on the sled in this distance?
J
(c) What is the work done by gravity on the sled?
J
(d) What is the total work done?
J
2-) A 575-g squirrel with a surface area of 915 cm2 falls from a 5.2-m tree to the ground. Estimate its terminal velocity. (Use the drag coefficient for a horizontal skydiver. Assume that the cross-sectional area of the squirrel can be approximated as a rectangle of width 11.4 cm and length 22.8 cm. Note, the squirrel may not reach terminal velocity by the time it hits the ground. Give the squirrel's terminal velocity, not it's velocity as it hits the ground.)
m/s
What will be the velocity of a 57.5-kg person hitting the ground, assuming no drag contribution in such a short distance?
m/s
Explanation / Answer
1.
Use below figure,
a) Wf= f*d = s*f*d = = s*mgcos*d = 0.1*95*9.8*cos66*30 = 1136 J
b) tension in the rope = T
by Newton’s second law ,
T+f - mgsin = ma
Since speed is constant a= 0 m/s^2
T = mgsin – f = mgsin - s*mgcos = mg(sin - s*cos)
Work done by T = work done by rope = Wt = T*d = mg(sin - s*cos)*d = 95*9.8*(sin66-0.1*cos66)*30 = 24379 J
c) Wg = - mgsin*d = 95*9.8*sin66*30 = -25515 J
d) Total work done = Wf+Wt+Wg = 1136 J + 24379 J - 25515 J = 0 J
2.
Ternimal velocity , V = sqrt[(2W)/(Cd**A)]
Where W= weight =mg, Cd= drag coefficient=1 , = density of air = 1.21 kg/m^3 , A= frontal area = 11.4 cm * 22.8 cm = 259.92 cm^2 = 0.02599 m^2
Plugging values,
V = sqrt[(2W)/(Cd**A)] = sqrt[(2*0.575*9.8)/(1*1.21*0.02599)] = 18.9 m/s
Velocity of the person , v = sqrt(2gh) = sqrt(2*9.8*5.2) = 10m/s
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