Question

20. From the following graph of an object of mass 20 kg, answer the questions that follow :

(a) Acceleration between 10-15 sec.
(b) What is the force applied from 0-5 sec ?​

Answer 1

Given:

• A velocity-time graph is given.

To Find:

• (a) Acceleration between 10-15 sec.
• (b) What is the force applied from 0-5 sec.

Answer:

We know that slope of a (v-t) graph gives acceleration.

So , slope = tan∅.

From 10 to 15 s base = (15-10) s = 5s.

Height = 20m/s - 10m/s = 10m/s.

= > acclⁿ = 10m/s ÷ 5s .

=> acclⁿ = 2m/s²

Hence the acclⁿ is 2m/s².

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For finding force applied we should know mass and acclⁿ.

Here already mass is given 2kg and we must now find acceleration.

From 0s to 5s :

Acclⁿ = slope = tan∅

Here base is (5-0)s = 5s.

Height is (10m/s - 0m/s) = 10m/s

=> acclⁿ = 10m/s÷5s = 2m/s².

We know Force as ,

★ FORCE = MASS × ACCLⁿ★

=> Force = 2kg × 2m/s².

=> Force = 4N.

Hence the force applied was 4N.

Answer 2

Answer:

(a) The acceleration is $2m/s^{2}$.

(b) The force applies is $0N$.

Explanation:

It is given that mass is $20kg$

(a) Acceleration between $10-15s$

From graph we can see that, velocity is $10m/s$ and $20m/s$.

So, the acceleration is

$a=\frac{20-10}{15-10}$

$a=\frac{10}{5}$

$a=2m/s^{2}$

(b) Time given is $0-5s$

We need to determine the force.

At this time, velocity is constant and constant velocity do not produce acceleration.

Acceleration is zero.

Hence, $F=ma$

Using this $F=0N$

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