P9: A 2.0 kg object moves along the x-axis under the influence of an unknown for
ID: 1778356 • Letter: P
Question
P9: A 2.0 kg object moves along the x-axis under the influence of an unknown force. Its 2.5 Inn' and b-12 Jima potential energy due to the forces U(x)s ar' + br', where a a) what is the force on the object when x = 0.5 m? b) If the object is released from rest at x = 0.5 m, what is its velocity when it crosses the origin? P10: An object of mass m slides from initial height h down a curved ramp, onto a flat area. There is a patch of length L with coefficient of kinetic friction ' on the flat portion, but the ramp is frictionless everywhere else. Beyond the patch, there is a spring of constant k. a) Assume friction is high enough to stop the mass half way across the patch what value of . will do this? Please be sure to answer in terms of the other given variables and g. b) Instead, assume that friction is low enough for the mass to reach the spring, how much is the spring compressed? Please be sure to answer in terms of the given variables and g. P11: A 2.5 kg block is dropped from 50 cm above a 500 N/m spring. You may ignore air resistance. 50 cm a) How far will the spring be compressed before the block comes to a momentary stop? b) The spring compresses, then launches the block back up into the air. What is the block's speed when it is 25 cm above the spring, heading back up? P12: A 2.0-kg block is held at rest against a spring of constant 2700 N/m, compressing it 55 cm. The block and the spring are located on a horizontal surface. When the system is released, the block slides down a ramp with an angle of 35e to a second horizontal surface 1.7 m below the first. The incline has a coefficient of kinetic .9 friction of 0.29; the horizontal surfaces are frictionless. Please find a) the potential energy stored in the spring before it is released. b) the mechanical energy lost by the block once it reaches the lower horizontal surface. c) the final speed of the block as it slides along the lower horizontal surfaceExplanation / Answer
P9]
The Potential Energy is given by: U(x) = ax4 + bx2
and Force is negative gradient of Potential Energy. So, to find the force, differentiate the potential energy with respect to x.
F = - [dU/dx]
=> F = - [4ax3 + 2bx]
now substitute for a and b to get F
F = - [4(2.5)(0.5)3 + 2(12)(0.5)] = - 13.25 N
b]
F = mdv/dt
so, dv/dt = a = - (10/m)x3 - (24/m)x
Now v2 = u2 - 2ax
=> v2 = -2x[-5x3 - 12x]
=> v2 = 2[0.5][5(0.5)3 + 12(0.5)]
=> v = 2.574 m/s
this is the velocity when it crosses the origin.
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