Question

A 200 kg car initially travels with a velocity of 10 m/s to the right. A net force F acts on the car for 5 s, which causes the velocity of the cart to increase to 30 m/s to the right.

Determine the magnitude of the force F
What direction does force F act on the car? - ¿

Answer 1

The force F acting on the 200 kg car, initially moving at 10 m/s, is 800 N to the right. The direction of the force aligns with the car's motion, causing acceleration.

To determine the magnitude of the force
`\( F \),` we can use Newton's second law of motion, which states:

`\[ F = m \cdot a \]`

where F is the force, m is the mass, and a is the acceleration. The acceleration a can be calculated using the equation of motion:

`\[ v_f = v_i + a \cdot t \]`

where:

-
`\( v_f \)` is the final velocity,

-
`\( v_i \)` is the initial velocity,

- \( a \) is the acceleration,

- \( t \) is the time.

Given that the initial velocity
`(\( v_i \))` is 10 m/s, the final velocity
`(\( v_f \))`is 30 m/s, and the time (\( t \)) is 5 s, we can rearrange the equation to solve for acceleration (\( a \)):

`\[ a = \frac{{v_f - v_i}}{{t}} \]`

Substitute the given values:

`\[ a = \frac{{30 \, \text{m/s} - 10 \, \text{m/s}}}{{5 \, \text{s}}} = \frac{{20 \, \text{m/s}}}{{5 \, \text{s}}} = 4 \, \text{m/s}^2 \]`

Now, use Newton's second law to find the force (\( F \)):

`\[ F = m \cdot a \]`

Given that the mass (\( m \)) is 200 kg and the acceleration (\( a \)) is 4 m/s²:

`\[ F = 200 \, \text{kg} \cdot 4 \, \text{m/s}^2 = 800 \, \text{N} \]`

So, the magnitude of the force \( F \) is 800 N.

As for the direction of the force, it acts in the same direction as the acceleration. In this case, since the car is initially moving to the right and the force causes the velocity to increase in the same direction, the force \( F \) acts to the right.