Question

A body of mass 5kg is moved up over 10m along the line of greatest slope of a smooth inclined plane of inclination of 30° with the horizontal if g=10m/s^2 then the work done will be

Answer 1

Answer:

Hey

Explanation:

The displacement against the gravitational force is Ssinθ

or

10sin30

o

=5m

Thus we get the work done as gravitational force x displacement (in vertical direction)

5g×5=250 J

Answer 2

Question :

• A body of mass 5kg is moved up over 10m along the line of greatest slope of a smooth inclined plane of inclination of 30° with the horizontal if g=10m/s^2 then the work done will be

Required Solution :

• The work done by the gravitational force is 250 Joules

Provided that :

• Mass of the body is 5 kg
• Displacement of the body is 10 m
• Angle of the inclined plane is 30°
• Acceleration due to gravity is 10 m/s²

To Find out :

• The work done on the body

Concept used :

• Formula to find the work done

Formula Used :

• Work done = Force × Displacement ( Cos ∅ )

Required knowledge :

»» The forces with act on a body when in inclined surface are mg Sin∅ towards the left side and mg Cos∅ perpendicularly to the block

»» The force required to move the the block will be equal to the component of force acting up on it which is mg Sin∅

We know that :

• Force applied = mg Sin∅
• Displacement = 10 m
• Angle between the force and displacement = 0

Full Solution :

$: \implies \sf Work \; Done = F \times S \times Cos \; \emptyset$

$: \implies \sf Work \; Done = Mg \; sin \emptyset \times 10 \times Cos ( 0 )$

$: \implies \sf Work \; Done = 5 \times 10 \times \sin ( 30 ) \times 10 \times 1$

$: \implies \sf Work \; Done = 5(10) \times \dfrac{1}{2} \times 10(1)$

$: \implies \sf Work \; Done = 50 \times \dfrac{1}{2} \times 10$

$: \implies \sf Work \; Done = 25 \times 10$

$: \implies \sf Work \; Done = 250 Joules$

• Henceforth the work done is 250 Joules