A cart of mass 20 kg at rest is to be dragged at a speed of 18 km h ?. If the co-efficient of friction
between the cart and the ground is 0.1, what is the minimum force required to drag the cart to a
distance of 10 m? (g = 10 m s?)
Answers
Given :
- mass of cart = 20 kg
- Speed = 18 km/h
- co-efficient of friction between the cart and the ground = 0.1
To find :
- Minimum force required to drag the cart to a distance of 10 m
Solution :
⟹ Speed = 18km/h
⟹ we know that 1 km = 1000m and 1 hour = 3600 seconds.
⟹ Therefore , (18×1000)/3600
⟹ 5 m/s
[ Refer attachment for free body diagram]
Now , we will find Kinetic energy :-
KE = 1/2 m v²
⟹ KE = 1/2 × 20 × (5)²
⟹ KE = 1/2 × 20 × 25
⟹ KE = 1 × 10 × 25
⟹ KE = 250 J
In free body diagram , fs = force due to friction and F = minimum force required to drag the cart to a distance of 10 m.
⟹ WD ( work done) = ∆ KE
WD = force × displacement
i.e F(10) -( µmg×10) = 250
( Now µ =0.1 )
⟹ F(10) -( 0.1 ×20 × 10×10) = 250
⟹ F(10) -( 1 ×20 × 10) = 250
⟹ F(10) -200 = 250
⟹ F(10) = 250+200
⟹ F(10) = 250+200
⟹ F× 10 = 450
⟹ F= 450/10
⟹ F= 45 N
Answer :
45 Newton force required to drag the cart to a
distance of 10m.
Given:
mass = 20 kg
speed = 18 km/h
co-efficient of friction = 0.1
Solution:
First, we have to change the units of given speed i.e,
18 km/h = 18 × 1000/3600
so, the speed = 5 m/s
Now, we have to find the kinetic enery of the cart
KE = 1/2 m × v^2
= 1/2 × 20 × (5)^2
= 10 × 25
= 250 J
:. kinetic energy = 250 Joules
So, the force of friction and Force is equal to the minimum force required to drag the cart to the distance of 10 m.
Work done = change in kinetic energy
WD = force × displacement i.e, F × (10) - (co-efficient of friction × mass × acceleration due to gravity × 10) = 250.
(we know that co-efficient of friction is 0.1)
=> F × (10) - (0.1 × 20 × 10) = 250
=> F × (10) - (1 × 20 × 10 × 10) = 250
=> F × (10) - 200 = 250
=> F × (10) = 250 + 200
=> F × (10) = 450
=> F = 450/10
=> F = 45 N
This is the required ans...
Hope it helped you...