A 1–kilogram block slides down a frictionless inclined plane 2.5 meters high. What is its velocity at the bottom of the plane? (Assume gravitational acceleration is 9.8 m/s2.)
A. 2.5 m/s
B. 3.9 m/s
C. 7.0 m/s
D. 9.8 m/s
Answers
answer : option (C) 7.0 m/s.
It is given that a 1–kilogram block slides down a frictionless inclined plane 2.5 meters high.
we have to fInd the velocity of the block at the bottom of the plane.
here m = 1kg, h = 2.5 m and g = 9.8 m/s²
using work - energy theorem,
energy at the highest point of the plane = energy at the bottom of the plane.
⇒ 1/2 m(0)² + mgh = 1/2 mv² + mg(0)
⇒mgh = 1/2 mv²
⇒v² = 2gh
⇒v² = 2 × 9.8 × 2.5
⇒v² = 9.8 × 5 = 49
⇒v = 7
therefore, velocity of the block at the bottom of the plane is 7 m/s.
Answer:
It is given that a 1–kilogram block slides down a frictionless inclined plane 2.5 meters high.
we have to fInd the velocity of the block at the bottom of the plane.
here m = 1kg, h = 2.5 m and g = 9.8 m/s²
using work - energy theorem,
energy at the highest point of the plane = energy at the bottom of the plane.
⇒ 1/2 m(0)² + mgh = 1/2 mv² + mg(0)
⇒mgh = 1/2 mv²
⇒v² = 2gh
⇒v² = 2 × 9.8 × 2.5
⇒v² = 9.8 × 5 = 49
⇒v = 7