A 10kg runaway grocery cart runs into a spring with spring constant 260 n/m and compresses it by 50 cm . Part A What was the speed of the cart just before it hit the spring?
1/2 m v^2=1/2 k x^2
\
k given, x=.5m
solve for v.
To find the speed of the cart just before it hit the spring, we can use the law of conservation of mechanical energy. The mechanical energy before and after the collision should remain constant.
The mechanical energy of an object is the sum of its kinetic energy (KE) and potential energy (PE). In this case, the potential energy is in the form of the elastic potential energy stored in the spring.
The kinetic energy of the cart just before it hits the spring can be calculated using the formula:
KE = (1/2) * mass * velocity^2
Where:
- KE is the kinetic energy
- mass is the mass of the cart (10 kg)
- velocity is the unknown speed of the cart
Next, we need to determine the potential energy stored in the compressed spring. The formula for potential energy in a spring is:
PE = (1/2) * k * x^2
Where:
- PE is the potential energy
- k is the spring constant (260 N/m)
- x is the compression of the spring (50 cm or 0.5 m)
According to the law of conservation of mechanical energy, the kinetic energy before the collision should be equal to the potential energy:
KE = PE
Substituting the values into the formulas:
(1/2) * mass * velocity^2 = (1/2) * k * x^2
Simplifying the equation:
10 * velocity^2 = 260 * (0.5)^2
10 * velocity^2 = 260 * 0.25
10 * velocity^2 = 65
Dividing both sides by 10:
velocity^2 = 6.5
Taking the square root of both sides:
velocity ≈ √6.5
Therefore, the approximate speed of the cart just before it hit the spring is √6.5 m/s.