Need help with these physics ques please?

A bowling ball of mass 6.7 kg and radius 9.0 cm rolls without slipping down a lane at 4.3 m/s. Calculate its total kinetic energy.

A 3.35 m long pole is balanced vertically on its tip. It starts to fall and its lower end does not slip. What will be the speed of the upper end of the pole just before it hits the ground? [Hint: Use conservation of energy.]
 
1)
KE = (1/2)(m)(v)^2 the 9.0 cm radius is not important for this
= (1/2)(6.7kg)(4.3 m/s)^2
= 61.9415 Joules
= 62 J rounded to the proper number of significant digits

2)
for the tip of the pole, using conservation of energy,
potential gravitational energy = kinetic energy at the bottom
ie. PE = KE
(m)(g)(change in height) = (1/2)(m)(v)^2
we can divide both sides by mass, 'm',
(g)(change in height) = (1/2)(v)^2
isolate the speed, "v",
2g(change in height) = v^2 multply both sides by 2
take the positive square root of both sides (cant have negative speed )
so the equation now becomes
square root of (2g(change in height)) = v
change in height = 3.35m
g = 9.8 m/s^2
input the values
square root of [2(9.8)(3.35)] = v
square root of 65.66 = v
8.1 m/s = v
therefore the speed of the pole just before it hits the ground is 8.1 m/s
 
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