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AP Physics 1: Block Acceleration, Velocity, and Distance

Created by @nathanedwards

at November 1st 2023, 5:08:50 pm.

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#acceleration

#ap-physics-1

#newtons-second-law

AP Physics 1 Exam Question:

A block of mass 2 kg is initially at rest on a frictionless horizontal surface. Suddenly, a net force of 10 N is applied horizontally to the block for a duration of 2 seconds. After this, the force is removed, and the block continues to slide.

a) Calculate the acceleration of the block during the 2 seconds period. b) Calculate the velocity of the block at the end of the 2 seconds period. c) Calculate the distance traveled by the block during the 2 seconds period.

Answer:

a) To calculate the acceleration of the block, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

The formula for Newton's second law is:

F_{\text{net}} = m \cdot a

Where:

$F_{\text{net}}$ is the net force acting on the block (10 N).
$m$ is the mass of the block (2 kg).
$a$ is the acceleration of the block (to be determined).

By rearranging the formula, we can solve for the acceleration:

a = \frac{F_{\text{net}}}{m}

Substituting the given values:

a = \frac{10 \, \text{N}}{2 \, \text{kg}} = 5 \, \text{m/s}^2

Therefore, the acceleration of the block during the 2 seconds period is 5 m/s^2.

b) To calculate the velocity of the block at the end of the 2 seconds period, we can use the formula:

v = u + a \cdot t

Where:

$v$ is the final velocity of the block.
$u$ is the initial velocity of the block (which is 0 m/s since the block starts at rest).
$a$ is the acceleration of the block (5 m/s^2).
$t$ is the time duration (2 seconds).

Substituting the given values:

v = 0 + 5 \, \text{m/s}^2 \cdot 2 \, \text{s} = 10 \, \text{m/s}

Therefore, the velocity of the block at the end of the 2 seconds period is 10 m/s.

c) To calculate the distance traveled by the block during the 2 seconds period, we can use the formula:

s = u \cdot t + \frac{1}{2} \cdot a \cdot t^2

Where:

$s$ is the distance traveled by the block.
$u$ is the initial velocity of the block (0 m/s).
$a$ is the acceleration of the block (5 m/s^2).
$t$ is the time duration (2 seconds).

Substituting the given values:

s = 0 \cdot 2 \, \text{s} + \frac{1}{2} \cdot 5 \, \text{m/s}^2 \cdot (2 \, \text{s})^2 = 10 \, \text{m}

Therefore, the distance traveled by the block during the 2 seconds period is 10 meters.