Question

Two horizontal forces act on a 4 kg disk that slides over frictionless ice. one force is in the positive x direction and has a magnitude of 7 N. The second force has a magnitude of 9 N. The x component of the disk's velocity changes from -4m/s to +5m/s in 3 seconds. calculate the direction of the second force.

Answer 1

acceleration is change of velocity per unit time. \[ a = \frac{v_f - v_i}{t} = \frac{5 m/s - (-4) m/s}{3 s} = 3 m/s^2 \] The force acting on the disk is \[F_x = m a = 4 kg \times 3 m/s^2 = 12 N\] That's the force in the x direction. We know that one force is in positive x direction with magnitude 7 N. The second force therefore has a component in x direction with magnitude 5 F. \[F_x = 12 N =F_{1,x} + F_{2, x} = 7 N + 5 N \]
We know that the second force has a magnitude of 9 N, its x component is 5 N. From trigonometry we know that \[F_x = F \cos(\varphi)\\ \Rightarrow \varphi = \arccos\left(\frac{F_x}{F}\right)\\ \Rightarrow \varphi = \arccos\left(\frac{5 N}{9 N}\right) = 0.981\]