a body of mass 5kg is moving with a momentum of 10 kg m/s. A force of 0.2 N acts on it in the direction of motion of body for 10 sec. The increase in kinetic energy?
Answers
Answer :
- The increase in kinetic energy of the body is 4.4 J.
Explanation :
Given :
- Mass of the body, m = 5 kg
- Momentum of the body, p = 10 kg m/s
- Force acting on the body, F = 0.2 N
- Time of contact of force, t = 10 s
To find :
- Increase in kinetic energy (K.E.), ∆K.E. = ?
Knowledge required :
Formula for momentum :
⠀⠀⠀⠀⠀⠀⠀⠀⠀p = mv
Where,
- p = Momentum of the body
- m = Mass of the body
- v = Velocity of the body
Formula for force :
⠀⠀⠀⠀⠀⠀⠀⠀⠀F = ma
Where,
- F = Force
- m = Mass
- a = Acceleration
First equation of motion :
⠀⠀⠀⠀⠀⠀⠀⠀⠀v = u + at
Where,
- v = Final velocity
- u = Initial velocity
- a = Acceleration
- t = Time Taken
Formula for Kinetic energy :
⠀⠀⠀⠀⠀⠀⠀⠀⠀K.E. = ½mv²
Where,
- K.E. = Kinetic energy
- m = Mass
- v = Velocity
Change in kinetic energy :
⠀⠀⠀⠀⠀⠀⠀⠀ ∆K.E. = K.E.' - K.E.
Where,
- ∆K.E. = Change in Kinetic energy
- K.E.' = Final kinetic energy
- K.E. = Initial Kinetic energy
Solution :
Orginal Kinetic energy :
First let us find the velocity of the body.
Using the equation for momentum and substituting the values in it, we get :
⠀⠀=> p = mv
⠀⠀=> 10 = 5 × v
⠀⠀=> 10/5 = v
⠀⠀=> 2 = v
⠀⠀⠀⠀⠀⠀⠀∴ v = 2 m/s
Hence the velocity of the ball is 2 m/s.
Now,
Let's find out the kinetic energy (Initial) of the body :
By using the Equation for Kinetic energy of a body and substituting the values in it, we get :
⠀⠀=> K.E. = ½mv²
⠀⠀=> K.E. = ½ × 5 × 2²
⠀⠀=> K.E. = ½ × 5 × 4
⠀⠀=> K.E. = 5 × 2
⠀⠀=> K.E. = 10
⠀⠀⠀⠀⠀⠀⠀⠀∴ K.E. = 10 J
Hence the initial kinetic energy possessed by the body is 10 J.
Final kinetic energy :
Let us find the acceleration of the body :
By using the equation for force and substituting the values in it, we get :
⠀⠀=> F = ma
⠀⠀=> 0.2 = 5 × a
⠀⠀=> 0.2/5 = a
⠀⠀=> 0.04 = a
⠀⠀⠀⠀⠀⠀⠀⠀∴ a = 0.04 m/s²
Hence the acceleration produced by the body is 0.04 m/s².
Now let us find the velocity of the body after 10s.
By using the first equation of motion and substituting the values in it, we get :
⠀⠀=> v = u + at
⠀⠀=> v = 2 + 0.04 × 10
[Note : Here the initial velocity is taken as 2 m/s as the body is starting from that velocity at t = 0 m/s]
⠀⠀=> v = 2 + 0.4
⠀⠀=> v = 2.4
⠀⠀⠀⠀⠀⠀⠀∴ v = 2.4 m/s
Hence the velocity of the body after 10 s is 2.4 m/s.
So now we can find the kinetic energy possessed by the body.
By using the Equation for Kinetic energy of a body and substituting the values in it, we get :
⠀⠀=> K.E. = ½mv²
⠀⠀=> K.E.' = ½ × 5 × 2.4²
⠀⠀=> K.E.' = ½ × 5 × 5.76
⠀⠀=> K.E.' = ½ × 28.8
⠀⠀=> K.E.' = 14.4
⠀⠀⠀⠀⠀⠀⠀⠀∴ K.E.' = 14.4 J
Hence the final kinetic energy possessed by the body is 14.4 J
To find the increase in kinetic energy of the body :
By using the equation for change in kinetic energy of the body and substituting the values in it, we get :
⠀⠀=> ∆K.E. = K.E.' - K.E.
⠀⠀=> ∆K.E. = 14.4 - 10
⠀⠀=> ∆K.E. = 4.4
Therefore,
- Increase in kinetic energy (K.E.), ∆K.E. = 4.4 J
Answer:
Answer :
The maximum hieght reached by the ball is 45 m.
Explanation :
Given :
Angle of projection, θ = 30°
Intial velocity of the ball, u = 60 m/s
Final velocity of the ball, v = 0 m/s
[Final velocity of a body at maximum height is 0, v = 0]
Acceleration due to gravity, g = 10 m/s² (Approx.)
[Acceleration due to gravity = 9.8 m/s²]
To find :
Maximum height reached by the ball, h = ?
Knowledge required :
Formula for maximum height in case of a projectile :
\boxed{\sf{h_{(max.)} = \dfrac{u^{2}sin^{2}\theta}{2g}}}
h
(max.)
=
2g
u
2
sin
2
θ
Where,
h = Maximum height
u = Initial velocity
θ = Angle of Projection
g = Acceleration due to gravity
Solution :
By using the formula for maximum height in case of a projectile and substituting the values in it, we get :
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{u^{2}sin^{2}\theta}{2g}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
2g
u
2
sin
2
θ
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{60^{2} \times sin^{2}30^{\circ}}{2 \times 10}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
2×10
60
2
×sin
2
30
∘
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{60^{2} \times \bigg(\dfrac{1}{2}\bigg)^{2}}{20}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
20
60
2
×(
2
1
)
2
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{60^{2} \times \dfrac{1}{4}}{20}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
20
60
2
×
4
1
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{3600 \times \dfrac{1}{4}}{20}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
20
3600×
4
1
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{3600}{20 \times 4}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
20×4
3600
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{3600}{80}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
80
3600
\begin{gathered}:\implies \sf{h_{(max.)} = \dfrac{360}{8}} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=
8
360
\begin{gathered}:\implies \sf{h_{(max.)} = 45} \\ \\ \\ \end{gathered}
:⟹h
(max.)
=45
\begin{gathered}\boxed{\therefore \sf{h_{(max.)} = 45\:m}} \\ \\ \\ \end{gathered}
∴h
(max.)
=45m
Therefore,
Maximum height reached by the ball, h = 45 m.