Question

dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The fric- tional force is negligible. The block starts from rest and moves 11.0 m in 5.00 s. (a) What is the mass of the block of ice? (b) If the worker stops pushing at the end of 5.00 s, how far does the block move in the next 5.00 s?

Answer 1

Answer:

Explanation:

Given

Force=80 N

Initial velocity(u)=0

Distance moved 11 m in 5 s

$s=ut+\frac{at^2}{2}$

$11=0+\frac{a\times 25}{2}$

$22=a\times 25$

$a=\frac{22}{25}$

$a=0.88 m/s^2$

and $force=mass\times acceleration$

$80=m\times 0.88$

m=90.9 kg

velocity acquired by block at the end of 5 sec

v=u+at

$v=0+0.88\times 5=4.4 m/s$

Distance moved in next 5 sec

s=vt+0

$s=4.4\times 5=22 m$

Answer 2

The constant horizontal force applies by dockworker on ice block is equal to the product of its mass and acceleration by, Newtons second law of motion .

• a) The mass of the block of ice is 90.9 kg.
• b) The distance traveled by the ice block in next five seconds is 22 meters.

What is Newtons second law of motion?

Newtons second law of motion shows the relation between the force mass and acceleration of a body.

It says, that the force applied on the body is equal to the product of mass of the body and the acceleration of it.

It can be given as,

$F=ma$

Here, (m) is the mass of the body and (a) is the acceleration.

Given information

The constant horizontal force applied by the dockworker to the block of ice is 80 N.

The frictional force is negligible.

The distance traveled by the block is 11 meters in 5 seconds.

• a) The mass of the block of ice-

As the distance traveled by the block is 11 meters in 5 seconds and the initial velocity of the block is zero. Thus by the distance formula of equation of motion, this distance can be given as,

$11=0+\dfrac{1}{2}a\times5^2\\a=0.88 \rm m/s^2$

Thus the acceleration of the block is 0.88 meter per squared second.

As the mass of the block is ratio of force applied on it to the acceleration of the block. Thus,

$m=\dfrac{80}{0.88}\\m=90.9\rm kg$

The mass of the block of ice is 90.9 kg.

• (b) The distance traveled by the ice block in next five seconds-

Final velocity of the block after the 5 seconds when the initial velocity is zero,

$v_f-0=at\\v_f=0.88\times5\\v_f=4.4 m/s$

As the distance traveled  by block is product of its velocity and the time taken by it. Thus the distance traveled by the ice block in next five seconds is,

$x=4.4\times5\\x=22\rm m$

Thus, the distance traveled by the ice block in next five seconds is 22 meters.

Hence,

• a) The mass of the block of ice is 90.9 kg.
• b) The distance traveled by the ice block in next five seconds is 22 meters.

Learn more about the Newtons second law of motion here;

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