Answer:
a) 607.5 J
b) 160.531875 J
c) 0 J
d) 0 J
e) 2.925 m\s
Explanation:
The given data :-
- Mass of the box ( m ) = 37.5 kg.
- Displacement made by box ( x ) = 4.05 m.
- Horizontal force ( F ) = 150 N.
- The co-efficient of friction between box and floor ( μ ) = 0.3
- Gravitational force ( N ) = m × g = 37.5 × 9.81 = 367.875
Solution:-
a) The work done by applied force ( W )
W = force applied × displacement = 150 × 4.05 = 607.5 J
b) The increase in internal energy in the box-floor system due to friction.
Frictional force ( f ) = μ × N = 0.3 × 367.875 = 110.3625 N
Change in internal energy = change in kinetic energy.
ΔU = ( K.E )₂ - ( K.E )₁
Since the initial velocity is zero so the ( K.E )₁ = 0
ΔU = ( K.E )₂ = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J
c) The work done by the normal force
.
Displacement of box vertically = 0
W = force applied × displacement = 367.875 × 0 = 0 J
d) The work done by the gravitational force.
Displacement of box vertically = 0
W = force applied × displacement = 367.875 × 0 = 0 J
e) The change in kinetic energy of the box
( K.E )₂ - ( K.E )₁ = ( K.E )₂ - 0 = ( F - f ) × ( x ) = ( 150 - 110.3625 ) × 4.05 = 160.531875 J
f) The final speed of the box
( K.E )₂ = 160.531875 J = 0.5 × 37.5 × v²
v² = 8.56
v = 2.925 m\s.
