How much kinetic energy does a 700-gram baseball have that is travelling at 20 m/s? What would be the kinetic energy if you doubled the mass? What would be the kinetic energy if you doubled the velocity?
V1= 20 m/s
K1= ½ m v1^2
= ½ (0.7 kg)(20 m/s)^2
= ½ (0.7 g)(400 m^2/ s^2)
= 140 J
Why did you not change to SCI units in the beginning instead of dividing by 1000 at the end?
double mass --> twice
double velocity --> 4 times
How much kinetic energy does a 700-gram baseball have that is travelling at 20 m/s? What would be the kinetic energy if you doubled the mass? What would be the kinetic energy if you doubled the velocity?
V1= 20 m/s
K1= ½ m v1^2
= ½ (700 g)(20 m/s)^2
= ½ (700 g)(400 m^2/ s^2)
= 140, 000/ 1,000
= 140 J
2. How much work must gravity do on a 2-kg falling object in order to take it from rest to 20 m/s?
W= KE
K= F*D
W= (Vf^2 -V1^2)/2)
W= ½ m (Vf^2-V1^2)
W= ½ m Vf^2 - ½ m V1^2
= ½ (2 kg)(20 m/s^2)
= 400 J
3. A 5-kg box is sliding across a desk that has a coefficient of friction of 0.4. If the box is initially moving at 15 m/s, how far will the box slide before coming to rest. You MUST use the WORK/Kinetic Energy Theorem to solve this. F= mu: F n = mu mg: N:
m= 5 kg V1= 15 m/s ÓFy= may
W= m (Vf^2 -V1^2 /2)
= ½ m (Vf^2 -V1^2)
= ½ m Vf^2 - ½ m V1^2
W= Kf - Ki
W= ÄK
3 answers
2. How much work must gravity do on a 2-kg falling object in order to take it from rest to 20 m/s?
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change in potential energy = m g h
that goes into kinetic energy.
so how far does it fall to go from 0 to 20 m/s?
v = Vi + a t
20 = 0 + 9.81 t
t = 2.04 seconds
how far in 2.04 s
h = (1/2)a t^2 = 4.9(2.04)^2
= 20.4 meters fall
so
m g h = 2 * 9.81 * 20.4 = 400 JOules so yes
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change in potential energy = m g h
that goes into kinetic energy.
so how far does it fall to go from 0 to 20 m/s?
v = Vi + a t
20 = 0 + 9.81 t
t = 2.04 seconds
how far in 2.04 s
h = (1/2)a t^2 = 4.9(2.04)^2
= 20.4 meters fall
so
m g h = 2 * 9.81 * 20.4 = 400 JOules so yes
3. A 5-kg box is sliding across a desk that has a coefficient of friction of 0.4. If the box is initially moving at 15 m/s, how far will the box slide before coming to rest. You MUST use the WORK/Kinetic Energy Theorem to solve this.
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initial ke = (1/2)(5)(225)
= 562.5 Joules
work done = mu m g x
= .4 (5)(9.81) x = 562.5
so
x = 28.7 meters
=============================
initial ke = (1/2)(5)(225)
= 562.5 Joules
work done = mu m g x
= .4 (5)(9.81) x = 562.5
so
x = 28.7 meters
5 answers