The increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.
<h3>What is gravitational potential energy?</h3>
The energy that an item has due to its location in a gravitational field is known as gravitational potential energy.
The potential energy increases by 3773 J
PE₂-PE₁=mg(h₂-h₁)
3773 J = 35.0 × 9.81 × (h₂-h₁)
(h₂-h₁) = 10.98
Case 2 ;
ΔPE =?
ΔPE=mg(h₂-h₁)
ΔPE=56.0 × 9.81 ×10.98
ΔPE=6031.97 J.
Hence, the increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.
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Answer:
4 m/s
Explanation:
KE=1/2mv^2
720=1/2(90)
720=45v^2
divide by 45
16= v^2
over the square root
4=v
v= 4m/s
Answer:
The chunk went as high as
2.32m above the valley floor
Explanation:
This type of collision between both ice is an example of inelastic collision, kinetic energy is conserved after the ice stuck together.
Applying the principle of energy conservation for the two ice we have based on the scenery
Momentum before impact = momentum after impact
M1U1+M2U2=(M1+M2)V
Given data
Mass of ice 1 M1= 5.20kg
Mass of ice 2 M2= 5.20kg
velocity of ice 1 before impact U1= 13.5 m/s
velocity of ice 2 before impact U2= 0m/s
Velocity of both ice after impact V=?
Inputting our data into the energy conservation formula to solve for V
5.2*13.5+5.2*0=(10.4)V
70.2+0=10.4V
V=70.2/10.4
V=6.75m/s
Therefore the common velocity of both ice is 6.75m/s
Now after impact the chunk slide up a hill to solve for the height it climbs
Let us use the equation of motion
v²=u²-2gh
The negative sign indicates that the chunk moved against gravity
And assuming g=9.81m/s
Initial velocity of the chunk u=0m/s
Substituting we have
6.75²= 0²-2*9.81*h
45.56=19.62h
h=45.56/19.62
h=2.32m
