A daredevil on a motorcycle leaves the end of a ramp with a speed of 34.9 m/s as
ID: 1603672 • Letter: A
Question
A daredevil on a motorcycle leaves the end of a ramp with a speed of 34.9 m/s as in the figure below. If his speed is 32.8 m/s when he reaches the peak of the path, what is the maximum height that he reaches? Ignore friction and air resistance. A 0.220-kg block along a horizontal track has a speed of 1.30 m/s immediately before colliding with a light spring of force constant 3.70 N/m located at the end of the track. (a) What is the spring's maximum compression if the track is frictionless? (b) If the track is not frictionless, would the spring's maximum compression be greater than, less than, or equal to the value obtained in part (a)? greater less equal A skier of mass 72 kg is pulled up a slope by a motor-driven cable. (a) How much work is required to pull him 85 m up a 30 degree slope (assumed frictionless) at a constant speed of 2.2 m/s? (b) What power (expressed in hp) must a motor have to perform this task?Explanation / Answer
In this case vertical or y direction, the velocity will be
=34.9sin45=24.6m/s
Atpeak velocity will be 32.8sin45=23.19m/s
Thus
V2=U2-2gh
23.19^2=24.6^2-2*9.8*h
h=(24.6^2-23.19)/(2*9.8)=3.43 m
====:::::
Applied force on spring isproportional to displacement or compression. Thus
F=kx
Here F=0.22*1.3 for a second interaction
k=3.7 N/mThus
x=F/k=(0.22*1.3)/3.7=0.077m
It will be less than for friction track.
========
Force required=72*9.8*85 =59976 N (independent of slope)
Power=force*velocity=59976*2.2=131947.2 watt
Time =85/2.2=38.63
Energy=power*time=131947.2*38.63=5093.162 kj
Power is already given divide by 786 to get in hp