Question

The graph in the figure shows the x component of the acceleration of a 2.4-kg object as a function of time (in ms).
(a) At what time(s) does the x component of the net force on the object reach its maximum magnitude?
(b) What is the maximum magnitude of the x component of the net force on the object?
(c) What is the x component of the net force on the object at time t = 0ms and at t = 4 ms?

Answer 1

Answer:

thanks for free points

Answer 2

Answer:

Acceleration of a 2.4-kg object as a function of time (in ms).a. $\text { the time is } \mathrm{t}=3 \mathrm{~ms}$ b.$\text { force }=-24 \mathrm{~N}$ c. $15m/s$

Explanation:

Step 1: A) The equation a = v/t denotes acceleration (a), which is the change in motion (v) over the change in time (t). This enables you to calculate the difference in velocity in meters per second squared $(m/s2).$

as the force =  acceleration * mass

the force is maximum when the acceleration is maximum

the maximum magnitude of acceleration ,$a = 20 m/s^2$

maximum magnitude of force$= 20 * 2.4$

maximum magnitude of force$= 48 N$

the time is$t = 3 ms$

Step 2: B) The formula below demonstrates how the mass of an item multiplied by its acceleration determines the size of the net force acting on it. An object is not accelerating and is in a condition known as equilibrium if the net force exerted on it is zero.

at t  $= 0 ms$

force $= 5 * 2.4$

force$= 12 N$

at t $= 4 ms$

force $= - 10 * 2.4$

force $= -24 N$

Step 3: C)

Position is given as $\mathrm{x}=\mathrm{a}+\mathrm{bt}^2=8.5+2.5 \mathrm{t}^2$

Position at t $=2 \mathrm{~s}, \mathrm{x}_2=8.5+2.5(2)^2=18.5 \mathrm{~m}$

Position at t $=4 \mathrm{~s}, \quad \mathrm{x}_1=8.5+2.5(4)^2=48.5 \mathrm{~m}$

Displacement  S $=\mathrm{x}_2-\mathrm{x}_1=48.5-18.5=30 \mathrm{~m}$

Time taken  t $=4-2=2 \mathrm{~s}$

Average velocity $\mathrm{V}_{\mathrm{avg}}=\frac{\mathrm{S}}{\mathrm{t}}=\frac{30}{2}=15 \mathrm{~m} / \mathrm{s}$

Hence acceleration of a 2.4-kg object as a function of time (in ms).a. $\text { the time is } \mathrm{t}=3 \mathrm{~ms}$ b.$\text { force }=-24 \mathrm{~N}$ c. $15m/s$

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