Kinetic energy comes when the object starts moving. This energy is due to motion. Although this energy is due to motion, this energy is not created. It is usually converted from one type of energy to another type.
Sample Problems
Question 1: A ball has a mass of 2 kg; suppose it travels at 10 m/s. Find the kinetic energy possessed by it.
Answer:
Given: m = 2Kg, and v = 10m/s
The KE is given by, K.E =
\frac{1}{2}mv^2 K.E =
\frac{1}{2}mv^2 ⇒ K.E =
\frac{1}{2}(2)(10)^2 ⇒ K.E = 100J
Question 2: A ball has a mass of 10 kg; suppose it travels at 100 m/s. Find the kinetic energy possessed by it.
Answer:
Given: m = 10Kg, and v = 100m/s
The KE is given by,
⇒ K.E =
\frac{1}{2}(10)(100)^2 ⇒ K.E = 50000J
Question 3: Work done by a force on a moving object is 100J. It was traveling at a speed of 2 m/s. Find the new speed of the object if the mass of the object is 2 kg.
Answer:
Given: W = 100J
Work done by the force is equal to the change in kinetic energy.
W =
\frac{1}{2}m(v^2 - u^2) Given, u = 2 m/s and v = ?, m = 2kg.
Plugging the values in the given equation,
W =
\frac{1}{2}m(v^2 - u^2) ⇒
100 = \frac{1}{2}(2)(v^2 - 2^2) ⇒
100 = v^2 - 2^2 \\ = 104 = v^2 \\ = v = \sqrt{104} \text{ m/s}
v = \sqrt{104} \text{ m/s}
Question 4: Work done by a force on a moving object is -50J. It was traveling at a speed of 10m/s. Find the new speed of the object if the mass of the object is 2 kg.
Answer:
Given: W = -50J
Work done by the force is equal to the change in kinetic energy.
W =
\frac{1}{2}m(v^2 - u^2) Given, u = 10m/s and v = ? . m = 2kg.
Plugging the values in the given equation,
W =
\frac{1}{2}m(v^2 - u^2) ⇒
-50 = \frac{1}{2}(2)(v^2 - 10^2) ⇒
-50 = v^2 - 10^2 \\ = 50 = v^2 \\ = v = \sqrt{50} \\ = v = 5\sqrt{2} \text{ m/s} The speed is decreased because the work done was negative. This means that the force was acting opposite to the block and velocity was decreased.
Question 5: Suppose a 1000 kg object was traveling at a speed of 10 m/s. Now, this mass transfers all its energy to a mass of 10 kg. What will be the velocity of the 10 kg mass after being hit by it?
Answer:
KE is given by the formula,
K.E =
\frac{1}{2}mv^2 KE of the heavier object
M = 1000Kg and v = 10m/s
K.E =
\frac{1}{2}mv^2 ⇒ K.E =
\frac{1}{2}(1000)(10)^2 ⇒K.E = 50,000J
Now this energy is transferred to another ball.
m = 10Kg and v = ?
50,000 =
\frac{1}{2}(10)v^2 ⇒ 10,000 = v2
⇒ v = 100 m/s
Question 6: Suppose a 10 kg mass was traveling at a speed of 100 m/s. Now, this mass transfers all its energy to a mass of 20 kg. What will be the velocity of the 20 kg mass after being hit by it?
Answer:
To find the velocity of the 20 kg mass after being hit by the 10 kg mass, we use the principle of conservation of momentum.
According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision, assuming no external forces are acting on the system.
m1 = 10kg
v1 = 100m/s
m2 =20kg
v2 = ?
The total momentum before the collision (Pinitial) is :
= m1 x v1 = 10 x 100 = 1000 Kgm/s
According to the conservation of momentum:
Pinitial = Pfinal
m1 x v1 = m2 x v2
10 x 100 = 20 x v2
1000 = 20 x v2
v2 = 1000/20 = 50m/s
So, the velocity of the 20 kg mass after being hit by the 10 kg mass is 50m/s.
Question 7: Suppose a 10 kg block was kept at a 20m height. Now, this block is dropped. Find out the velocity of the block just before it hits the ground.
Answer:
The block of 10Kg is kept at a height of 20m.
The potential energy of the block will be,
P.E = mgh
Here m = 10, g = 10m/s2 and h = 20m.
P.E = mgh
⇒ P.E = (10)(10)(20)
⇒ P.E = 2000J
Now, this energy is converted completely into KE.
KE = PE
⇒2000 =
\frac{1}{2}mv^2 Given m = 10Kg,
⇒2000 =
\frac{1}{2}10v^2 ⇒400 = v2
v = 20m/s
Question 8: Suppose a rock of 100 kg was kept at an 80m height. Now, this block is dropped. Find out the velocity of the block just before it hits the ground.
Answer:
The block of 10Kg is kept at a height of 80m.
The potential energy of the block will be,
P.E = mgh
Here m = 100, g = 10m/s2 and h = 80m.
P.E = mgh
⇒ P.E = (100)(10)(80)
⇒ P.E = 80000J
Now, this energy is converted completely into KE.
KE = PE
⇒80000 =
\frac{1}{2}mv^2 Given m = 100Kg,
⇒80000 =
\frac{1}{2}100v^2 ⇒1600 = v2
v = 40m/s