Question

A Jeep of mass 2000kg is moving with a velocity of 54km/h hits a wall and comes to rest in 10s. Find the force exerted by the jeep on the wall. Need answer quick!

Answer 1

Answer:

• Force exerted by jeep on the wall is -3000 N.

Step-by-step explanation:

Given :

• Mass of jeep (m) = 2000 kg
• Initial velocity of jeep (u) = 54 km/h
• Final velocity of jeep (v) = 0
• Time taken (t) = 10 s

To Find :

• Force exerted by jeep on the wall (F)?

Solution :

Firstly converting units of initial velocity from km/h to m/s,

⇒ Initial velocity (u) = 54 × 5/18

⇒ Initial velocity (u) = 270/18

⇒ Initial velocity (u) = 15 m/s

Now finding its acceleration by using first equation of motion. As we know that,

• v = u + at

Where,

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time taken

We have,

• Final velocity (v) = 0 m/s
• Initial velocity (u) = 15 m/s
• Time taken (t) = 10 s
• We have to find acceleration

Putting all values,

⇒ 0 = 15 + a(10)

⇒ 0 = 15 + 10a

⇒ 15 + 10a = 0

⇒ 10a = 0 - 15

⇒ 10a = -15

⇒ a = -15/10

⇒ a = -1.5 m/s²

• Hence, acceleration of jeep is -1.5 m/s².

Now finding force exerted by jeep on the wall. As we know that,

• F = ma

Where,

• F denotes force exerted
• m denotes mass
• a denotes acceleration

We have,

• Mass (m) = 2000 kg
• Acceleration (a) = -1.5 m/s²
• We have to find force exerted (F)

Putting all values,

⇒ F = (2000)(-1.5)

⇒ F = 2000 × -1.5

⇒ F = -3000 N

• Hence, force exerted by jeep on the wall is -3000 N.

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Answer 2

Refer to the attachment