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Can someone please help me with these two they go together to make one question

Can someone please help me with these two they go together to make one question-example-1
User Fazil
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1 Answer

5 votes

1) Since the truck is moving, it has kintic energy. The formula for calculating kinetic energy is expressed as

KE = 1/2mv^2

where

m = mass

v = velocity

From the information given,

m = 60000

v = 27

Thus,

Kinetic energy = 1/2 x 60000 x 27^2 = 21870000 J

The amount of work that must be done to stop the truck must be equal to or greater than the kinetic energy. Thus,

amount of work that must be done to stop the truck = 21870000 J

2) Net force applied = - 300000 N

Recall,

Net Force = mass x acceleration

acceleration = net force/mass

acceleration of the truck = - 300000 /60000 = - 5 m/s^2

Since the acceleration is negative, it is decelerating.

To find the distance before it stops, we would apply one of Newton's formula of motion which is expressed as

v^2 = u^2 + 2as

where

v = final velocity

u = initial velocity

a = acceleration

s = distance

From the information given,

v = 0(becuase it will stop moving)

u = 27

a = - 5

By substituting these values into the formula,

0^2 = 27^2 - 2 x 5 x s

0 = 729 - 10s

10s = 729

s = 729/10

s = 72.9

The truck will move 72.9 meters before stopping

User Resnbl
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