Dynamics: Question 2

Worked solution

Part (a): Weight of the robot

Weight is the gravitational force on a mass, $W = mg$: $W = 12 \times 9.81 = 117.72\text{ N}$

So the weight of the robot is $\boxed{118}\text{ N}$ (to 3 s.f.).

Part (b): Resultant horizontal force

Two horizontal forces act on the robot: the $30\text{ N}$ driving force forward, and the $6.0\text{ N}$ resistive force backward (opposing motion). Since these act in opposite directions, the resultant is their difference: $F_{\text{net}} = 30 - 6.0 = 24\text{ N}$

The resultant force is $\boxed{24}\text{ N}$, directed forward (in the direction of travel), since the driving force is larger than the resistive force.

Part (c): Acceleration from Newton’s second law

Newton’s second law states $F_{\text{net}} = ma$, so: $a = \frac{F_{\text{net}}}{m} = \frac{24}{12} = 2.0\text{ m s}^{-2}$

The acceleration is $\boxed{2.0}\text{ m s}^{-2}$, in the same direction as the resultant force, forward, in the direction of travel.

Part (d): Acceleration once the driving force is switched off

With the driving force removed, the only horizontal force remaining is the $6.0\text{ N}$ resistive force, which acts backward (opposing the direction of travel). Applying $F=ma$ with this single force: $a' = \frac{6.0}{12} = 0.5\text{ m s}^{-2}$

Since the only remaining force acts backward, this acceleration is directed backward, opposite to the robot’s direction of travel, the robot decelerates at $0.5\text{ m s}^{-2}$.

Consistency check: the new acceleration ($0.5\text{ m s}^{-2}$) is exactly a quarter of the original ($2.0\text{ m s}^{-2}$), which makes sense since the net force driving the motion has dropped from $24\text{ N}$ to $6.0\text{ N}$, a factor of $4$, while the mass is unchanged.

Final answers

• (a) Weight $= \boxed{118}\text{ N}$
• (b) Resultant force $= \boxed{24}\text{ N}$, forward
• (c) Acceleration $= \boxed{2.0}\text{ m s}^{-2}$, forward
• (d) New acceleration $= \boxed{0.5}\text{ m s}^{-2}$, backward (a deceleration)