Question

The tow spring on a car has a spring constant of 3,086 N / m and is initially stretched 18.00 cm by a 100.0 kg college student on a skateboard. Which of the following is the velocity when the potential energy is 20.0 J?
Select one:

a. This problem cannot be solved without knowing the time because velocity is a function of time.

b. 0.774 m / s

c. 1.00 m / s

d. 0.600 m / s

Answer 1

Answer:

The velocity of the skateboard is 0.774 m/s.

Explanation:

Given that,

The spring constant of the spring, k = 3086 N/m

The spring is stretched 18 cm or 0.18 m

Mass of the student, m = 100 kg

Potential energy of the spring, $P_f=20\ J$

To find,

The velocity of the car.

Solution,

It is a case of conservation of energy. The total energy of the system remains conserved. So,

$P_i=K_f+P_f$

$\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2+20$

$\dfrac{1}{2}\times 3086\times (0.18)^2=\dfrac{1}{2}mv^2+20$

$50-20=\dfrac{1}{2}mv^2$

$30=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{60}{100}}$

v = 0.774 m/s

So, the velocity of the skateboard is 0.774 m/s.