Suppose you apply a force of 75 N to a 25-kg object. What will the acceleration of the object b? (Remember a=F/m)

Physics

1 answer:

Vlada [557]2 years ago

8 0

I would choose the option B.

F = ma

a = 75 / 25 = 3 m/s^2

You might be interested in

What is potential energy called that is associated with objects that can be stretched

Umnica [9.8K]

Elastic potential energy

5 0

2 years ago

A 77.0 kg firefighter climbs a flight of stairs 8.0 m high. how much work is required? 0 incorrect: your answer is incorrect. j

Alex73 [517]

<span>6160 joules

to lift 1 newton 1 metre requires 1 joule
there are 10 newtons in one kilo

so 77(kg) x 8 (metres) x 10 (newtons/kilo) = 6160 joules</span>

5 0

3 years ago

a 800 kg roller coaster cart is accelerated by a constant net force over a distance of 10 meters, as shown in the graph below: d

devlian [24]

Answer:

vf = 22.36[m/s]

Explanation:

First we must understand the data given in the problem:

m = mass = 800 [kg]

F = force = 20000[N]

dx = displacement = 10[m]

From newton's second we know that the sum of forces must be equal to the product of mass by acceleration.

$F = m*a\\20000 = 800*a\\a = 20000/800\\a = 25 [m/s^2]$

With the calculated acceleration, we can use the kinematics equations.

$v_{f} ^{2} =v_{o} ^{2}+2*a*dx\\ v_{o} = initial velocity = 0\\a = acceleration = 25[m/s^2]\\dx= displacement = 10[m]\\$

The key to using this equation is to clarify that the initial velocity is zero since the body is at rest, otherwise the initial velocity would be an initial data.

$v_{f} =\sqrt{2*25*10} \\v_{f} =22.36[m/s]$

Another way of solving this problem is by means of the definition of work and kinetic energy, where work is defined as the product of the force by the distance.

W =F*d

W = 20000*10

W = 200000[J]

Kinetic energy is equal to work, therefore the value calculated above is equal to:

$E_{k}=W =0.5*m*v_{f}^{2} \\200000=0.5*800*v_{f}^{2}\\v_{f}=\sqrt{\frac{200000}{0.5*800} } \\v_{f}=22.36[m/s]$