Calculating Energy Types
A 0.5 kg bird flies at 2 m/s 50 meters above a canyon below. What is the bird's gravitational potential energy?
GPE = mgh
= (0.5)(9.8)(50)
= 245 J
A 5 kg block sits at rest on the ground. What is the block's total mechanical energy?
0 J
A pencil gets dropped from a 1 meter high desk. The pencil does not move to start. What is the speed of the pencil just before it hits the ground?
GPE0 = KE
mgh0 = 1/2mv2
gh0 = 1/2v2
(9.8)(1) = 1/2v2
9.8 = 1/2v2
19.6 = v2
v = 4.4 m/s
What is the work/energy theorem in equation form?
W= Δ E
A 0.5 kg bird flies at 2 m/s 50 meters above a canyon below. What is the bird's kinetic energy?
KE = 1/2mv2
= 1/2 (0.5)(2)2
= 1 J
A 65 kg diver (which happens to be the mass of Tom Holland) stands on the edge of a pool 1 meter above the water below. What is the diver's total mechanical energy relative to the water below?
TME = GPE
TME = mgh
TME = (65)(9.8)(1)
TME = 637 J
A pencil rolls of a 1 meter high desk at a speed of 2 m/s. What is the speed of the pencil just before it hits the ground?
GPE0 + KE0 = KE
mgh0 + 1/2mv02= 1/2mv2
gh0 +1/2v02= 1/2v2
(9.8)(1) + 1/2(2)2 = 1/2v2
9.8 + 1/2(4) = 1/2v2
9.8 + 2 = 1/2v2
11.8 = 1/2v2
23.6 = v2
v = 4.9 m/s
A basketball has 15 J of energy to start, and then a player pushes the ball in the opposite direction it was moving in, doing 5 J of work on the ball. What is the basketball's final energy?
E0+Wenv=E
15 + -5 = E
E = 10 J
A spring of spring constant 300 N/m stretches 10 cm. What is the potential energy stored in the spring?
SPE = 1/2kx2
= 1/2(300)(0.1)2
= 1/2(300)(0.01)
= 1.5 J
A 65 kg diver (which happens to be the mass of Tom Holland) runs at 3 m/s before jumping into a pool 1 meter above the water below. What is the diver's total mechanical energy relative to the water below?
TME = KE + PE
= KE + GPE
= 1/2 (65)(3)2 + (65)(9.8)(1)
= 1/2 (65)(9) + 637
= 292.5 + 637
= 929.5 J
A feather of mass 6 grams drops from rest off a 1 meter high desk. If 0.05 J of energy are dissipated from friction of the feather with the air, what is the feather's speed just before it hits the ground?
GPE0 = ENM + KE
mgh0 = ENM + 1/2mv2
(0.006)(9.8)(1) = 0.05 + 1/2(0.006)v2
0.0058 = (0.003)v2
2.93 = v2
v = 1.7 m/s
A person pushes a 8 kg box, initially at rest, across a frictionless surface for a distance of 3 meters. If the person applied an average force of 10 N on the box over this distance, what is the change in energy of the box?
W = Δ E = Fdcosθ
= (10)(3)cos0
= +30 J
A moving bird initially has 2 J of kinetic energy as it glides through the air. Suddenly, the bird sees a fish and swoops down, tripling its speed. What is the bird's new kinetic energy?
v x 3
v and KE have a square relationship, so
KE x (3)2
KE x (9)
2 x 9 = 18 J
A 65 kg person runs at 1 m/s across a diving board of elasticity constant 600 N/m which bends 0.5 meters downward toward the water below. The height of the board above the water when it is not bent is 5 meters. What is the person's total mechanical energy before the dive?
TME = PE + KE
= mgh + 1/2kx2 + 1/2mv2
= (65)(9.8)(5-0.5) + 1/2(600)(0.5)2 + 1/2(65)(1)2
= (65)(9.8)(4.5) + 1/2(600)(0.25) + 1/2(65)(1)
= 2866.5 + 75 + 32.5
= 2974 J
A feather of mass 6 grams slides off a 1 meter high desk at a speed of 1 m/s. If 0.05 J of energy are dissipated from friction of the feather with the air, what is the feather's speed just before it hits the ground?
GPE0 + KE0 = ENM + KE
mgh0 + 1/2mv2 = ENM + 1/2mv2
(0.006)(9.8)(1) + 1/2(0.006)(1)2 = 0.05 + 1/2(0.006)v2
0.0588 + 0.003 = 0.05 + (0.003)v2
0.0618 = 0.05 + (0.003)v2
0.0118 = (0.003)v2
3.9 = v2
v = 2.0 m/s
A person pushes an 8 kg box, initially at rest, across a frictionless surface for a distance of 3 meters. If the person applied an average force of 10 N on the box over this distance, what is the final speed of the box?
E0 + Wenv = E = KE since all the energy of the box in the end is KE.
Wenv = Δ E = Fdcosθ
= (10)(3)cos0
= +30 J
E0 = 0 J since the box started from rest
E0 + Wenv = E = KE
0 + 30 J = KE
KE = 1/2mv2
30 = 1/2(8)v2
60 = 4v2
15 = v2
v = 3.9 m/s
A spring is initially stretched 2 cm and stores 5 J of SPE. How many joules of energy does it store when stretched 5 cm?
First, we need to determine the factor of change of the spring stretch, ie what we multiple the original stretch with to get the new stretch. We can solve the equation below to determine this number.
2x = 5
x = 2.5
So,
x*2.5
x and SPE in the SPE equation have a square relationship, so as a result,
SPE x (2.5)2
SPE x 6.25
5 J x 6.25 = 31.25 J
An unstretched spring of constant 750 N/m dangles a total of 5 meters above the ground below. When 1 kg mass is hung on the spring, the spring stretches a total of 10 cm. What is the total mechanical energy of the mass if it is at rest when it stretches the spring 10 cm?Use g = 10 m/s/s.
TME = PE + KE
= GPE + SPE
= mgh + 1/2 kx2
= (1)(10)(5 m - 0.1 m) + 1/2(750)(0.1)2
= (1)(10)(4.9) + 1/2(750)(.01)
= 49 + 3.75
= 52.75 J
A 5 kg block at rest compresses a spring of spring constant 2000 N/m a total of 10 cm. The block then slides up a rough hill, creating 3 J of heat. At what height on the hill does the block end up?
SPE0 = ENM + GPE
1/2kx02 = ENM + mgh
1/2(2000)(0.12) = 3 + (5)(9.8)h
1/2(2000)(0.01) = 3 + 49h
10 = 3 + 49h
7 = 49h
h = 0.14 meters, or 14 cm
A person pushes a 6 kg box, initially at rest, across a surface for a distance of 4 meters. The surface exerts a force of friction on the box of 8 N. If the person applied an average force of 12 N on the box over this distance, what is the final speed of the box?
E0 + Wenv = E
E0 + Wperson + Wfriction = E = KE since all the energy of the box in the end is KE.
Wperson = Fpersondcosθ
= (12)(4)cos0
= +48 J
Wfriction = Ffrictiondcosθ
= (8)(4)cos180
= -32 J
E0 = 0 J since the box started from rest
E0 + Wperson + Wfriction = E = KE
0 + 48 - 32 = KE
KE = 16 J
KE = 1/2mv2
16 = 1/2(6)v2
32 = 6v2
5.3 = v2
v = 2.3 m/s