Question

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Answer 1

Question :

A cart of mass 20 kg at rest is to be dragged at a speed of 18 km/hr. If the coefficient of friction between the cart and the ground is 0.1, what is the minimum force required to drag the cart to a distance of 10 m? (Take g = 10 m/s²)

Answer :

Given -

• Mass of the cart = 20 kg
• Speed = 18 km/hr
• The coefficient of friction between cart and the ground = 0.1
• Distance = 10 m
• g = 10 m/s

To Find -

• The minimum force ?

Solution -

First of all, converting the units of speed given :

➸ 18 km/hr

➸ 5/18 * 18 m/s

➸ 5 m/s

Now, Finding the acceleration.

➸ v² - u² = 2as

➸ 5² - 0² = 2*a*10

➸ 25 = 20*a

➸ a = 25/20

➸ Acceleration = 1.25 m/s²

Now, To Find total minimum net force, we need to consider the force of friction along with the net force.

That is, F = umg + ma.

➸ Force = 0.1*20*10 + 20*1.25

➸ Force = 20 + 25

➸ Force = 45 N

Hence, the minimum force required to drag the cart to a distance of 10 m is 45 N.

Answer 2

»Question -

A cart of mass 20 kg at rest isto be dragged at a speed of 18km/h. If the co- efficient of friction between the cart and ground is 0.1, what is the minimum force required to drag cart to distance of 10m ? (take g = 10 m/s²)

» Given :

m = 20 kg

u = 0

v = 18 km/h

${\mu (coefficient \:of \:friction)= 0.1}$

s = 10m

» We know that,

${\pink{\boxed{F = ma}}}$

${\blue{\boxed{f (frictional \:force)= \mu N}}}$

${\green{\boxed{N (normal\:force)= mg}}}$

» To find acceleration -

${We\:know \:that,}$

v² = u²+2as

first we will convert 18 km/h into m/s

${\boxed{1 km/h = 5/18 m/s}}$

18 ×5/18

= 5

18 km/h = 5 m/s

by putting values

(5)² = (0)² +2a(10)

25 = 20a

a = 25/20

a = 1.25 m/s²

» To find normal force :

N = mg

N = 20 ×10

= 200 N

» Total force = Frictional force + ma

= ${\mu N+ ma}$

by putting values

(0.1 × 200) +(20 × 1.25)

= 20 +25

= 45 N

therefore,

${\pink{\boxed{Total\: force= 45N}}}$