Suppose a 94.5 kg mass is 3.75 m above the ground, and is dropped from rest. a)

Question

Suppose a 94.5 kg mass is 3.75 m above the ground, and is dropped from rest.

a) If there is no friction, at what speed will the mass hit the ground?
m/s

b) Suppose the mass now hits the ground at speed 5.72 m/s. How much work was done by friction on the mass? (Give the answer as a positive number.)
J

2. In procedure 1: a box of mass M starts from rest at height h, and slides down a frictionless incline.

a) If the mass of the box increases, the acceleration of the box on the incline
decreases
increases
remains the same


b) Suppose the speed of M at the bottom of the incline was v. At what height would M have to start so the speed of M at the bottom is 2v?

4h
h
between h and 2h
between 2h and 4h
more than 4h
2h

3. In procedure 2: an inclined plane is 120 cm long, and is tipped at angle ? = 10.5o. The 0 cm mark is at the bottom. Photogate 1 is at the 29.3 cm mark, and photogate 2 is at the 62.5 cm mark along the track, Assume

- The bottom of the track is at 0 J potential energy.
- The total mass of the cart is 654 g.
- The track is totally frictionless

NOTE: Be careful of units!


a) Find the potential energy of the cart .
- At photogate 1: J
- At photogate 2: J

b) Suppose the cart is released at rest at photogate 2 (the higher point on the track). Find the speed of the cart as it passes photogate 1.
m/s

4. –in procedure 3: a spring is at the bottom of a frictionless incline. When a box of mass M compresses the spring distance x and is released from rest, the box slides distance d up the incline (measured from where the box is at rest).

a) Suppose a box of mass 2M is used to compress the spring distance x,and is released from rest. The distance the box will slide up the incline is

d/2
between d and 2d
d/4
2d
between d/2 and d/4
between d/2 and d
d
4d
between 2d and 4d


b) Suppose a box of mass M is used to compress the spring distance 2x, and is released from rest. The distance the box will slide up the incline now is

d/2
2d
between d and 2d
between d and d/2
d/4
4d
between d/4 and d/2
between 4d and 2d
d

Explanation / Answer

a) From the kinematic relations v^2 - u^2 = 2gs Here u =0 g =9.80 m/s h = 3.75 m The speed will the mass hit the ground is v= sqrt[ 2(9.80 m/s2)(3.75 m)] = 8.573214 m/s b) From the law of conservation of energy mgh = (1/2)mv^2+ Wf The work was done by the friction is Wf = mgh- (1/2)mv^2 = (94.5 kg)(9.80 m/s2)(3.75 m) -(1/2)(94.5 kg )( 5.72 m/s)^2 = 1926.9306 J Note: Please submit one question per post

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