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ollegr [7]
4 years ago
8

A 1000kg truck traveling at 108km/h skids 10m before it stops. What is the magnitude of the frictional force acting on the car?

Physics
1 answer:
Leokris [45]4 years ago
3 0

Given that,

Mass of truck = 1000 kg

Speed = 108 km/h = 30 m/s

Distance = 10 m

We need to calculate the acceleration

Using equation of motion

v^2=u^2+2as

Where, v = final speed

u = initial speed

s = distance

Put the value into the formula

(0)^2=30^2+2\times a\times 10

a=-\dfrac{30^2}{2\times10}

a= -45 m/s^2

We need to calculate the magnitude of the frictional force acting on the truck

Using formula of force

F=ma

Where, m = mass of truck

a = acceleration

Put the value into the formula

F=1000\times(-45)

F= -45000\ N

Negative sign shows the opposite direction of motion.

Hence, The magnitude of the frictional force acting on the truck is 45000 N.

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Answer:

<h2>6000 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 3000 × 2

We have the final answer as

<h3>6000 N</h3>

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Complete question:

A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa

Answer:

The average pressure exerted on the surface by the block is 9655.17 Pa

Explanation:

Given;

density of the lead, ρ =  1.13 x 10⁴ kg/m³

mass of the lead block, m = 20 kg

surface area of the area of the block, A = 2.03 x 10⁻² m²

Determine the force exerted on the surface by the block due to its weight;

F = mg

F = 20 x 9.8

F = 196 N

Determine the pressure exerted on the surface by the block

P = F / A

where;

P is the pressure

P = 196 / (2.03 x 10⁻²)

P = 9655.17 N/m²

P = 9655.17 Pa

Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa

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