Practice 1. A 0.50 kg laboratory dynamics cart with an initial velocity of 3m / s [right] accelerates for 2.0 s at 1.2m / (s ^ 2) [right] when pulled by a string. Assume there is no friction acting on the cart.
(a) Calculate the force exerted by the string on the cart. [ans: 0.60 N]

From Newton's second law of motion, the net force on an object is equal to its mass times its acceleration (∑F = ma). Drawing a free body diagram of the cart, the only horizontal force is a tension force T pulling to the right.

∑F = ma

T = ma

T = (0.050 kg) (1.2 m/s²)

T = 0.060 N

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Practice 1. A 0.50 kg laboratory dynamics cart with an initial velocity of 3m / s [right] accelerates for 2.0 s at 1.2m / (s ^ 2) [right] when pulled by a string. Assume there is no friction acting on the cart. (a) Calculate the force exerted by the string on the cart. [ans: 0.60 N] ​

Answer :

Other Questions

Practice 1. A 0.50 kg laboratory dynamics cart with an initial velocity of 3m / s [right] accelerates for 2.0 s at 1.2m / (s ^ 2) [right] when pulled by a string. Assume there is no friction acting on the cart.
(a) Calculate the force exerted by the string on the cart. [ans: 0.60 N]