1) A 0.26-kg rock is thrown vertically upward from the top of a cliff that is 29
ID: 3898050 • Letter: 1
Question
1) A 0.26-kg rock is thrown vertically upward from the top of a cliff that is 29 m high. When it hits the ground at the base of the cliff the rock has a speed of 27 m/s.
(a) Assuming that air resistance can be ignored, find the initial speed of the rock.
(b) Find the greatest height of the rock as measured from the base of the cliff.
2) A block of mass m is attached to a string that is wrapped around the circumference of a wheel of radius R and moment of inertia I. The wheel rotates freely about its axis and the string wraps around its circumference without slipping (see Example 10-6). Suppose the block has a mass of 2.1 kg and an initial upward speed of 0.43 m/s. Find the moment of inertia of the wheel if its radius is 8.0 cm and the block rises to a height of 6.1 cm before momentarily coming to rest.
Explanation / Answer
1 ) Throw the stone up with 27 m/s, and you will see, how high it was:
height = v^2/(2g)
height = 27 ^2/(2*9,81) = 37.19 m.
This was 37.19 - 29 = 8.19 m above the cliff.
And again: h = v0^2/(2g):
v0^2 = 2gh
v0^2 = 2*9,81*8.19
v0 = 12.6 m/s --> = velocity up on the cliff
b)Vf^2=Vi^2+2ad, Vi=0m/s,
(27)^2=(0)^2+2(9.81)d
625=19.62d
d = 37.1 m
2) Final KE = 0 Final P.E = mgh = 2.1*9.8*0.061 =1.255 J
= 2*1.255/(5.375)^2 = 0.0868kg*m^2
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.