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Murljashka [212]
3 years ago
12

Your cat Goldie (mass 8.50 kg ) is trying to make it to the top of a frictionless ramp 39.930 m long and inclined 27.0 ∘ above t

he horizontal. Since Goldie can’t get any traction on the ramp, you push her up the entire length of the ramp by exerting a constant 37.40 N force parallel to the ramp (there is a set of stairs alongside the frictionless ramp on which you can walk). If Goldie is moving at 2.000 m/s at the bottom of the ramp (assume she has gotten a running start), what is her speed when she reaches the top of the incline?
Physics
1 answer:
Citrus2011 [14]3 years ago
7 0

Answer:

The speed when se reaches the top of the incline is 0.28 m/s

Explanation:

The work done is equal to the change of kinetic energy, then:

Wg + Wf + Wn = ΔEk

Where

Wg = work done by gravity

Wf = work done by force

Wn = work done by normal force

-mgdsin\theta +Fd+0=\frac{1}{2} *m*(v_{2} ^{2} -v_{f} )

Where

m = 8.5 kg

g = 9.8 m/s²

d = 39.93 m

F = 37.4 N

vf = 2 m/s

Replacing:

39.93*(-8.5*9.8*sin27+34.4)=\frac{1}{2} *8.5*(v_{2} ^{2} -2^{2} )\\v_{2} =0.28m/s

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Answer:

v (minimum speed) = 2.90 m/sec.

\\ \\ maximum speed (v)= 6.57 m/sec.\\

Maximum value of speed will occur at lowest point of vertical circle.

Explanation:

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Using the force balance expression at the top of the circle,

Gravitational Force + Tension force = Centrifugal force

m*g + T = m*v^2/R

Given that : T = 0

R = length of string = 0.86 m

mass of the spinning rock = 0.75 kg

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b) what is the maximum speed the rock can have so that the string does not break?

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T = m*g + m*v^2/R

Given that:

maximum tension T = 45 N

maximum speed v = ??

mass  m = 0.75 kg

∴

45 - 0.75*9.81 = 0.75*\frac{v^2}{0.86} \\\\v^2 = 0.86*(45 - 0.75*9.81)/0.75 \\ v = \sqrt{0.86*(45 - 0.75*9.81)/0.75\\ maximum speed (v)= 6.57 m/sec.\\

c)

At what point in the vertical circle does this maximum value occur?

Maximum value of speed will occur at lowest point of vertical circle.

This is so because  at the lowest point; the tension in string will be maximum.

4 0
2 years ago
if a 0.040-kg stone is whirled horizontally on the end of a 60-m string at a speed of 4.4 m/s, what is the centripetal force ? *
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evablogger [386]

Answer:

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If you have the following options:

<u>A.  1.63</u>

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