A 5 kg box slides down a long, frictionless
incline of angle 32. It starts from rest at time
t = 0 at the top of the incline at a height 20 m
above the ground. Find the potential energy of the box at
t = 1 s .
Answer in units of J
incline of angle 32. It starts from rest at time
t = 0 at the top of the incline at a height 20 m
above the ground. Find the potential energy of the box at
t = 1 s .
Answer in units of J
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The force that is accelerating the box will be the component of its weight parallel to the slope, which is given by
F = mg sin (theta) = 5 * 9.8 * sin (32) = 25.97 N
Its acceleration will then be a = F/m = 25.97/5 = 5.2 m/s/s
The distance it will travel in 1 second is given by d = ut + 1/2 a t^2
Since u = 0 the first term goes to 0. The second term is 1/2 * 5.2 * 1^2 = 2.6 m.
This is the distance travelled down the slope, so you'll need some trig to calculate the vertical distance travelled. It ends up being 2.6 sin 32 = 1.38 m.
Now, the object's potential energy is given by its height above the floor. It started at 20 and dropped 1.38 m, so it is now 20-1.38 = 18.62 m high
PE = mgh = 5 * 9.8 * 18.62 = 912.38 J
F = mg sin (theta) = 5 * 9.8 * sin (32) = 25.97 N
Its acceleration will then be a = F/m = 25.97/5 = 5.2 m/s/s
The distance it will travel in 1 second is given by d = ut + 1/2 a t^2
Since u = 0 the first term goes to 0. The second term is 1/2 * 5.2 * 1^2 = 2.6 m.
This is the distance travelled down the slope, so you'll need some trig to calculate the vertical distance travelled. It ends up being 2.6 sin 32 = 1.38 m.
Now, the object's potential energy is given by its height above the floor. It started at 20 and dropped 1.38 m, so it is now 20-1.38 = 18.62 m high
PE = mgh = 5 * 9.8 * 18.62 = 912.38 J
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http://www.ajdesigner.com/phppotentialen…
http://www.ajdesigner.com/phppotentialen…