A block with mass m = 2.42 kg is placed against a spring on a frictionless incli

Question

A block with mass m = 2.42 kg is placed against a spring on a frictionless incline with angle ? = 24.5° (see the figure). (The block is not attached to the spring.) The spring, with spring constant k = 25 N/cm, is compressed 29.8 cm and then released. (a) What is the elastic potential energy of the compressed spring? (b) What is the change in the gravitational potential energy of the block-Earth system as the block moves from the release point to its highest point on the incline? (c) How far along the incline is the highest point from the release point?

Explanation / Answer

Here,

theta = 24.5 degree

k = 25 N/cm = 2500 N/m

m = 2.42 Kg

x = 29.8 cm

a)

elastic potential energy of compressed spring = 0.5 * k * x^2

elastic potential energy of compressed spring = 0.5 *2500 * 0.298^2

elastic potential energy of compressed spring = 111 J

the elastic potential energy of compressed spring is 111 J

b)

change in gravitational potential energy of the block = initial stored energy in spring

change in gravitational potential energy of the block = 111 J

the change in gravitational potential energy of the block is 111 J

c)

let the distance is d

Now ,

m * d * g * sin(theta) = change in gravitational potential energy of the block

2.42 * 9.8 * d * sin(24.5) = 111

solving for d

d = 11.3 m

the block will travel 11.3 along the incline

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.