Answer to Question #113306 in Mechanics | Relativity for Maryam

Question

A 2.0 kg object is moving across a table at constant speed of 2.0 m/s against a force of friction of 6.0 N. The applied force is suddenly increased by 1.0 N. How fast will the object be moving after 3.0 s?

Answer 1

Explanation

• First, the object was moving at a constant velocity meaning that no acceleration along the moving direction ; hence applied force & the frictional force are equal in magnitude.
• After the sudden increment of the applied force,a net unbalanced force of (7N-6N =) 1N is being applied on the object resulting an acceleration & increasing its velocity thereafter.
• Soon after the motion turns to a linear motion under an acceleration with an initial velocity of 2ms^-1.

Notations

• Refer to the sketch attached.

Calculations

• Apply F = ma to the object along the moving direction

$\qquad\qquad \begin{aligned} \small 1N &= \small 2kg \times a_1\\ \small a_1 &= \small 0.5ms^{-2} \end{aligned}$

• Finding the velocity after 3s can be found by linear motion just started under the new acceleration.

$\qquad\qquad \begin{aligned} v &= u+at\\ v &= \small 2ms^{-1} + 0.5ms^{-2} \times 3s \\ v &= \small \bold{3.5ms^{-1}} \end{aligned}$

Good Luck