1. A student expends energy at the rate of 175 joules per second while lifting their backpack 1.3 meters in .75 seconds. Calculate how much work the student does lifting their backpack. Calculate the mass of the backpack. (Answer: 131.3 J , 10.3 kg)
2. A 1.2 gram feather is released from rest from a height of 2.1 meters. The feather hits the ground at 1.5 m/s. Calculate how much mechanical energy is lost. Calculate the average force of friction applied to the feather. (Answer: .223 J, .111N)
3. A 1500 kg car traveling at 30m/s hits a tree and comes to rest. The tree doesn't move and the front end of the car collapses at .65 meters. Calculate the loss of mechanical energy. Calculate the average force applied to the car. (Answer: 6.76 x 10^5 J , 1.04 x 10^6 N)
I have the answers but I need to know how to work these problems.
Please help! My teacher tends to get off topic so I wasn't able to learn how to do these, and I have a test tomorrow!
2. A 1.2 gram feather is released from rest from a height of 2.1 meters. The feather hits the ground at 1.5 m/s. Calculate how much mechanical energy is lost. Calculate the average force of friction applied to the feather. (Answer: .223 J, .111N)
3. A 1500 kg car traveling at 30m/s hits a tree and comes to rest. The tree doesn't move and the front end of the car collapses at .65 meters. Calculate the loss of mechanical energy. Calculate the average force applied to the car. (Answer: 6.76 x 10^5 J , 1.04 x 10^6 N)
I have the answers but I need to know how to work these problems.
Please help! My teacher tends to get off topic so I wasn't able to learn how to do these, and I have a test tomorrow!
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1.
W = P*t
W = 175*.75 = 131.3 J
W = F*d
F = W/d = m*g
m = W/ (g*d) = 131.3 / (9.81*1.3) = 10.3 kg
2.
In a vacuum, the feather would fall without resistance. It's potential energy is:
Ep = m*g*h = .0012*9.81*2.1 = 24.7*10^-3 J
It's kinetic energy on hitting the ground is:
Ek = (1/2)*m*v^2 = (1/2)*.0012*(1.5)^2 = 1.35*10^-3 j
The difference is the energy lost:
ΔE = 24.7*10^-3 - 1.35*10^-3 = .0234 J
This ΔE is equal to the work done or:
ΔE = F*d
F = ΔE/d = .0234/2.1 = .0118 N
Is there a typo in the question transcription?
3.
At speed, the kinetic energy is:
Ek = (1/2)*m*v^2 = (1/2)*1500*(30^2) = 675*10^3 J
Since the car comes to rest, the entire kinetic energy is lost, or 675*10^3 J
The lost energy is equal to the work done in decelerating the car:
W = F*d = Ek
F = Ek/d = 675*10^3 / .65 = 1.04*10^6 N
W = P*t
W = 175*.75 = 131.3 J
W = F*d
F = W/d = m*g
m = W/ (g*d) = 131.3 / (9.81*1.3) = 10.3 kg
2.
In a vacuum, the feather would fall without resistance. It's potential energy is:
Ep = m*g*h = .0012*9.81*2.1 = 24.7*10^-3 J
It's kinetic energy on hitting the ground is:
Ek = (1/2)*m*v^2 = (1/2)*.0012*(1.5)^2 = 1.35*10^-3 j
The difference is the energy lost:
ΔE = 24.7*10^-3 - 1.35*10^-3 = .0234 J
This ΔE is equal to the work done or:
ΔE = F*d
F = ΔE/d = .0234/2.1 = .0118 N
Is there a typo in the question transcription?
3.
At speed, the kinetic energy is:
Ek = (1/2)*m*v^2 = (1/2)*1500*(30^2) = 675*10^3 J
Since the car comes to rest, the entire kinetic energy is lost, or 675*10^3 J
The lost energy is equal to the work done in decelerating the car:
W = F*d = Ek
F = Ek/d = 675*10^3 / .65 = 1.04*10^6 N
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LOL I have the test tomorrow too :P
I figured out number one...
P=175W
S=.75s
d=1.3m
to find work, use the Power equation, P=w/t and manipulate it--> Pt=W
(175W)(.75s)=131.1J
That's your potential energy, so use the Potential energy equation to find mass
PE=mgh --> m=PE/gh
m=(131.1J)/(9.8m/s^2)(1.31m)
m=10.31kg
Number 2 I keep on getting .0223J :u idk why...
I figured out number one...
P=175W
S=.75s
d=1.3m
to find work, use the Power equation, P=w/t and manipulate it--> Pt=W
(175W)(.75s)=131.1J
That's your potential energy, so use the Potential energy equation to find mass
PE=mgh --> m=PE/gh
m=(131.1J)/(9.8m/s^2)(1.31m)
m=10.31kg
Number 2 I keep on getting .0223J :u idk why...