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andrew-mc [135]
3 years ago
14

How much force does it take to bring a 1,050 N car from rest to a velocity of 42 m/s in 13 seconds?

Physics
1 answer:
frozen [14]3 years ago
5 0

Answer:

F = 339.23 N

Explanation:

Weight of a car, W = 1050 N

Initial velocity, u = 0

Final velocity, v = 42 m/s

Time, t = 13 s

The weight of an object is given by :

W = mg

g is the acceleration due to gravity

m=\dfrac{W}{g}\\\\m=\dfrac{1050}{10}\\\\m=105\ kg

The force acting on car is :

F=ma\\\\F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{105\times (42-0)}{13}\\\\F=339.23\ N

So, the force acting on the car is 339.23 N.

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2 years ago
A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at
yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

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Therefore, the work done by the worker in lifting the bucket is given as:

W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)

Now, plug in the values given and solve for 'W'. This gives,

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