Respuesta :
The question is incomplete! The complete question along with answer and explanation is provided below.
Question:
A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.
What is the change in the potential energy (in Joules) of the mass as it goes up the incline?
If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?
Given Information:
Mass = m = 0.5 kg
Horizontal distance = d = 40 cm = 0.4 m
Vertical distance = h = 7 cm = 0.07 m
Normal force = Fn = 1 N
Required Information:
Potential energy = PE = ?
Work done = W = ?
Answer:
Potential energy = 0.343 Joules
Work done = 0.39 N.m
Explanation:
The potential energy is given by
PE = mgh
where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.
PE = 0.5*9.8*0.07
PE = 0.343 Joules
As you can see in the attached image
sinθ = opposite/hypotenuse
sinθ = 0.07/0.4
θ = sin⁻¹(0.07/0.4)
θ = 10.078°
The horizontal component of the normal force is given by
Fx = Fncos(θ)
Fx = 1*cos(10.078)
Fx = 0.984 N
Work done is given by
W = Fxd
where d is the horizontal distance
W = 0.984*0.4
W = 0.39 N.m
The potential energy of the mass is 0.343 J
The work done by the given force is 0.4 J
The complete question is given below:
A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.
What is the change in the potential energy (in Joules) of the mass as it goes up the incline?
If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?
Given information:
Mass m = 0.5 kg
length of the inclined surface l = 40 cm = 0.4 m
height of the top of the inclined plane h = 7 cm = 0.07 m
Normal force N = 1 N
Work and Energy:
The potential energy is given by
PE = mgh
where m is the mass of the object, h is the height
PE = 0.5×9.8×0.07
PE = 0.343 J
Work done by the force F = 1N parallel to the surface is given by
W = Fxd
where d is the horizontal distance
W = 1×0.4
W = 0.4 J
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