Question

a block of mass 50kg is pulled up a slope with a constant speed by applying an external force of 20kg wt. parallel to the slope.The block moves through a distance of 4m along the slope.
(i) What is the work done by the external applied force?
(ii) Name the S. I. unit of work. Express it in kg, metre, and sec. ​

Answer 1

Answer:

ANSWER

W=F×s

Here angle between force and displacement is 0 degree.Force applied due to weight of body is zero as in this case gravitational force and displacement are perpendicular.

F=150N;S=5 m.

W=150×5=750J

P.E=m×g×h

m×g=200N;h=3 m because one of the perpendicular side has length of 3 m

P.E=200×3=600J

Hence difference is 750−600=150J.

Answer 2

Answer:

i) 800J

ii) joule

(kg m²)/s²

Explanation:

Given that:

force = 20kg wt.= 20kgf = (20 × 10)N

displacement = 4m

Here,

It is given that force is parallel to displacement therefore force and displacement are in same direction hence positive work done thus cos theta=1.

Now we know that:-

i) work done = force × displacement × cos theta

= (20 × 10 × 4 × 1) J

= 800J ---(i)(ans)

ii) Express it in kg, metre, and sec :-

work done = force × displacement

= (kg m)/s² × m

= (kg m²)/s² ---(ii) (ans)