Your job is to lift 30 kgkg crates a vertical distance of 0.90 mm from the ground onto the bed of a truck. For related problem-s
1 answer:
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Correct question:
Consider the motion of a 4.00-kg particle that moves with potential energy given by
a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?
b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?
Answer:
a) 3.33 m/s
b) 0.016 N
Explanation:
a) given:
V = 3.00 m/s
x1 = 1.00 m
x = 5.00
At x = 1.00 m
= 4J
Kinetic energy = (1/2)mv²
= 18J
Total energy will be =
4J + 18J = 22J
At x = 5
= -0.24J
Kinetic energy =
= 2Vf²
Total energy =
2Vf² - 0.024
Using conservation of energy,
Initial total energy = final total energy
22 = 2Vf² - 0.24
Vf² = (22+0.24) / 2
= 3.33 m/s
b) magnitude of force when x = 5.0m
At x = 5.0 m
= 0.016N