Question

5 boxes are being pulled by 9.0 × 10^3 Newton force the mass of each box is 1.0 into 10 ^ 3 kg calculate the value of the acceleration produced if the force of friction in negligible.

Answer 1

Answer:

Acceleration = 1.8ms^-2

Explanation:

Data :

F = 9 × 10^3 N

m (mass) of each box = 10^3 kg

m of 5 boxes = 5 × 10^3 kg

Now, F = ma

Therefore, a = F/m

Substitute the values

We get, a = 9 × 10^3 ÷ 5 × 10^3

Cancel the 10^3 from both numerator and denominator.

We get a = 9 ÷ 5 = 1.8ms^-2

Therefore, your answer is 1.8ms^-2.

Please correct me if I'm wrong.

Answer 2

Concept

The push or pull on a mass-containing item changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude. Your instruction at this level should place a strong emphasis on the notion that a push or a pull is referred to as a force. There are several instances of forces in daily life, including: heft and force (i.e. the weight of something) the force a bat applies to a ball. the pressure that a hair brush applies to hair when brushing it.

Given

Five boxes are being pulled by $9.0 * 10^3 Newton$ force the mass of each box is $1.0$ into $10 ^ 3 kg$.

To Find

We have to find the value of acceleration.

Solution

According to the problem,

Total mass = $5*10^3kg$.

Force =  $9.0 * 10^3 Newton$

Acceleration = Force / mass = $\frac{9*10^3}{5*10^3} =9/5=1.8m/s^2$

Hence, acceleration is $1.8m/s^2$.

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