Question

At a waterpark, sleds with riders are sent along a slip-pery, horizontal surface by the release of a large compressed spring. The spring, with force constant k = 40.0 N/cm and negli-gible mass, rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 70.0 kg arc pushed against the other end, compressing the spring 0.375 m. The sled is then released with zero initial velocity. What is the sled's speed when the spring (a) returns to its uncompressed length and (b) is still compressed 0.200 m?

Answer 1

Answer:

a) 2.8m/s

b) 2.4m/s

Explanation:

Given:

K= 40N/cm, convert to N/m we have: k= 4000N/m

m=70kg

x=0.375

For (a)

We find the energy stored in the spring with the expression:

$U= 1/2 kx^2$;

$U= 1/2 * 4000 * 0.375^2$

U = 281 N.m

To find the sled speed, let's use the work energy theorem which is:

$V = \sqrt{(U*2)/m}$

Therefore

$V = \sqrt{(281*2)/70}$

V= 2.8m/s

b) We need to calculate energy stored at 0.200m.

Let's use:

$U = 1/2 kx^2$;

$U = 1/2 * 4000 * (0.375^2 - 0.2^2)$

U = 201N.m

We now use energy theorem:

We now have

$V = \sqrt{(201*2)/70}$

$V= \sqrt{402/70}$

V = 2.4m/s

Answer 2

Answer:

(a) v = 2.83m/s

(a) v = 2.40m/s

Explanation:

(a) This problem involves the concept of conservation of of mechanical energy

E = ΔPE = ΔKE

The elastic potential energy of the spring is converted into the kinetic energy of the sled. The potential energy is constant as there is no change in height of the sled and passenger.

So

ΔPE = 1/2×kx²

ΔKE = 1/2×mv²

The sled starts from rest

1/2×kx² = 1/2×mv²

1/2 is common to both sides

kx² = mv²

v² = kx²/m

v = √(kx²/m)

Given that k = 40.0N/cm

= 40.0×100N/m = 4000N/m

x = 0.375m

m = 70.0kg

v = √(4000×0.375²/70)

v = 2.83m/s

(b) ΔPE = 1/2×k(x1² ‐ x2²)

ΔKE = 1/2×mv²

1/2×mv² = 1/2×k(x1² ‐ x2²)

v² = k(x1² ‐ x2²)/m

v² = √(k(x1² ‐ x2²)/m)

v² = √(4000(0.375² ‐ 0.200²)/70)

v = 2.40m/s