Answer to Question #169853 in Physics for shawn

Question

A 50-kg sack of rice, resting on the ground, is dragged by a 100-N force at 30 degrees with the horizontal. What is the acceleration of the sack?

Answer 1

I assume there is no friction in this problem.

The horizontal component of the $F = 100N$ force is:

$$F_x = F\cos\theta$$

where $\theta = 30\degree.$

According to the second Newton's law, the acceleration provided by the net force acting on a body (in horizontal direction) is:

$$a = \dfrac{F_x}{m} = \dfrac{F\cos\theta}{m}$$

where $m = 50kg$ is the mass of the sack. Thus, obtain:

$$a =\dfrac{100\cos30\degree}{50} = \sqrt{3}\space m/s^2\approx 1.73\space m/s^2$$

Answer. $\sqrt{3}\space m/s^2\space \text{or}\space 1.73\space m/s^2$