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3 years ago

What acceleration will result when a 12 N net force applied to a 3kg object? A 6kg object?

Physics

1 answer:

olganol [36]3 years ago

8 0

As per Newton's II law we know that

$F_{net} = ma$

here for the first case we know that

$F_{net} = 12 N$

m = 3 kg

now we will have

$12 = 3\times a$

$a = 4 m/s^2$

now for other case we will have

m = 6 kg

now we have

$12 = 6 \times a$

$a = 2 m/s^2$

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A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on

MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

$Ec=\frac{1}{2} *m*v^{2}$

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

$W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}$

$W=\frac{1}{2} *m*(v2^{2}-v1^{2})$

In this case:

W=?
m= 2,145 kg
v2= 12 $\frac{m}{s}$

v1= 25 $\frac{m}{s}$

Replacing:

$W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})$

W= -515,872.5 J

The work required is -515,872.5 J

3 0

3 years ago

Calculate the work done (in J) by a 90.0 kg man who pushes a crate 4.25 m up along a ramp that makes an angle of 20.0° with the

Mrrafil [7]

Answer:

W = 3.4x0³ J.

Explanation:

The work done by the man is given by the following equation:

$W = F_{t}\cdot d$ (1)

where W: is the work, Ft is the total force and d: is the displacement = 4.25 m.

We need to find first the total force Ft, which is:

$Ft = Fm + W$

where Fm: is the force exerted by the man = 535 N, W: is the weight = m*g*sin(θ), m: is the mass of the man, g: is the gravitational acceleration = 9.81 m/s², and θ: is the angle = 20.0°.

$F_{t} = Fm + W = 500 N + 90.0 kg*9.81 m/s^{2} * sin(20.0) = 802.0 N$