Question

A small block of mass 200 g is kept at the top of a frictionless incline which is 10 m long and 3.2 m high. How much work was required (a) to lift the block from the ground and put it at the top, (b) to slide the block up the incline? What will be the speed of the block when it reaches the ground if (c) it falls off the incline and drops vertically on the ground (d) it slides down the incline? Take g=10 m/s².
Concept of Physics - 1 , HC VERMA , Chapter " Work and Energy"

Answer 1

Given in the question :-
Mass of the block = 200 g = 0.2 kg
Length of Incline, I = 10 m.
Height of the frictionless incline, h = 3.2 m

(a) Now required work done is equal to work done which against the gravity.
W = mgh
W =  0.2 × 10 × 3.2
W = 6.40 J.

(b) Since Here the incline is frictionless so the only force which work here is gravity. 
Hence in this also the work done against the gravity is same .
 W= mgh  = 6.40 J

(c) When the block falls on the ground vertically then :- It's speed
v² = u² + 2gh
v² = 0 + 2 × 10 × 3.2
v² = 64
v = 8 m/s

(d) When it reaches down the ground by sliding :-

Here only force working on the block is gravity , So the change in Kinetic energy is equal to work done by gravity.
∴ Change in Kinetic energy = Work done by gravity 
  mgh = 1/2 (mv²)
   v² = 2gh
   v² = √2gh
   v ²= √2 × 10 × 3.2
   v² = √64
   v = 8 m/s.
Hence it slide down the incline with velocity 8 m/s


Hope it Helps.

Answer 2

HEY!!

_______________________________

✴Mass of the block, m = 200g = 0.2kg

✴Length of the incline, s = 10m

✴Height of the incline, h = 3.2m

✴Acceleration due to gravity = 10m/sec^2

a)
✔✔Work done, W = mgh = 0.2 × 10 × 3.2 = 6.4J


b) Work done to slide the block up the incline

✔✔W = (mg sin θ) × s= (0.2) ×10 ×(3.2/10 )×10 =6.4 J

c) Let final velocity be v when the block falls to the ground vertically.

⚫Change in the kinetic energy = Work done

⚫12mv2−0=6.4 J

✔✔ν=8 m/s

d) Let ν be the final velocity of the block when it reaches the ground by sliding.

⚫12mν2−0=6.4 J

✔✔v=8 m/s