1. You do a physics lab experiment on another planet. A small block is released
ID: 1329123 • Letter: 1
Question
1. You do a physics lab experiment on another planet. A small block is released from rest at the top of a long frictionless ramp that is inclined at an angle of 36.9° above the horizontal. You measure that a small block travels a distance 16.0 m down the incline in 8.20 s. What is the value of g, the acceleration due to gravity on this planet
2.(a) A small block is released from rest at a height of 1.53 m above the ground and moves downward in free fall.
(i) How long does it take the block to reach the ground?
(ii) What is the speed of the block just before it strikes the ground?
(b) A frictionless incline is 5.00 m long (the distance from the top of the incline to the bottom, measured along the incline). The vertical distance from the top of the incline to the bottom is 1.53 m. A small block is released from rest at the top of the incline and slides down the incline.
(i) How long does it take the block to reach the ground?
(ii) What is the speed of the block just before it strikes the ground?
Explanation / Answer
1. by drawing a diagram
Angle =36.9 deg
The block has the mass m. Start from rest
Distance s=16 m at t= 8.2 sec
The acc. = a and g=g'
The net force Fnet = mg' *sinA
Newton's law: m*a = mg' * sinA
g' = a/sinA
v = v0 + a*t = 0 + a*t
s = at^2/2
16 = a*(8.2)^2/2
a = 32/(8.20^2) = 0.475
g' = a/sin = 0.475/sin(36.9) = 0.791 m/sec^2
2. A(ii). v^2 = u^2 + 2as
v^2 = 0^2 + 2(9.8)(1.53)
v^2 = 29.98
v = 5.47
A(i). v= u + a*t
5.47 = 0 + 9.8*t
t = 5.47/9.8 = 0.558 sec
B(i) a = g*sinA, on a inclined plane
sin A = 1.53/5.00 = 0.306
a = 9.8*0.306 = 2.9988 = 3.00 m/sec^2
s = u*t + 0.5*a*t^2
5 = 0*t + 0.5*3*t^2
t^2 = 5/1.5 = 1.82 sec.
B(ii) v^2 = u^2 + 2*a*s
v^2 = 0 + 2*3*5
v^2 = 30 = 5.47 m/sec^2
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