1. A small block travels up a frictionless incline that is at an angle of 30.0°
ID: 1329068 • Letter: 1
Question
1. A small block travels up a frictionless incline that is at an angle of 30.0° above the horizontal. The block has speed 4.19 m/s at the bottom of the incline. Assume g = 9.80 m/s2. How far up the incline (measured parallel to the surface of the incline) does the block travel before it starts to slide back down?
2.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?
3.Complete the following exercises. (Assume
g = 9.80 m/s2.)
(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?
(c) How do the answers in part (b) compare to the answers in part (a)?
(Select all that apply.
Speeds are considered approximately equal if they are within 5% of each other.)
It takes the block in (a) longer to reach the ground.The speed of the block in (a) is greater.It takes the block in (b) longer to reach the ground.The speed of the block in (b) is greater.The speeds are approximately equal.
Explanation / Answer
1)
Let h is height reached from the bottom in vertical direction.(Not along the incline)
Apply, energy conservation
K1 + U1 = K2 + U2
0.5*m*v^2 + 0 = 0 + m*g*h
h = 0.5*v^2/g
= 0.5*4.19^2/9.8
= 0.896 m
distance travelled along the incline, d = L/sin(30)
= 0.896/0.5
= 1.79 m <<<<<<<<<---------Answer
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