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Norma-Jean [14]
3 years ago
5

A horizontal force of 150 N is used to push a 38.0 kg packing crate a distance of 6.85 m on a rough horizontal surface. If the c

rate moves at constant speed, find each of the following.
a) find the work done by the 150N force:
b) find the coefficient of kinetic energy between the crate and rough surface
Physics
1 answer:
SIZIF [17.4K]3 years ago
3 0

Answer:

0.40

Explanation:

a) Work done is expressed as

W = Force × distance

Given.

Force = 150N

Distance = 6.85m

Workdone = 150×6.85

Work done = 1027.5Joules

b) According to Newton's second law

\sumFx = ma

Fm - Ff = ma

Since speed is constant, acceleration us zero

Fm - Ff = 0

Fm = Ff = nR

Fm is the moving force

Ff is the frictional force

n is the coefficient of kinetic energy between the crate and rough surface

R is the reaction

From the formula;

Fm = nR

n = Fm/R

n = Fm/mg

n = 150/38(9.8)

n = 150/372.4

n = 0.40

Hence the coefficient of kinetic energy between the crate and rough surface is 0.40

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{ \boxed{ \bold{ \sf{Acceleration \: ( \: a) = 8 \: m/ {s \: }^{2} }}}}

▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁▁

♨ Question :

  • A car starts from rest and is moving at 60.0 m/s after 7.50 s. What is the car's average acceleration ?

♨ \underbrace{ \sf{Required \: Answer : }}

☄ Given :

  • Initial velocity ( u ) = 0
  • Final velocity ( v ) = 60.0 m/s
  • Time ( t ) = 7.50 s

☄ To find :

  • Acceleration ( a )

✒ We know ,

\boxed{ \underline{ \bold{ \sf{Acceleration \: ( \: a) =  \frac{Final velocity ( v )  - Initial velocity ( u)}{t} }}}}

Substitute the values and solve for a.

➛ \sf{a =  \frac{60.0 - 0}{7.50}}

➛ \sf{a =  \frac{60.0}{7.50}}

➛ \boxed{ \boxed{ \sf{a = 8 \: m/ {s \: }^{2} }}}

---------------------------------------------------------------

✑ Additional Info :

  • When a certain object comes in motion from rest , in the case , initial velocity ( u ) = 0
  • When a moving object comes in rest , in the case , final velocity ( v ) = 0
  • If the object is moving with uniform velocity , in the case , u = v.
  • If any object is thrown vertically upwards in the case , a = -g
  • When an object is falling from certain height , in the case , final velocity at maximum height ( v ) = 0.

Hope I helped!

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