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What is the acceleration on a 2kg object that had 400J of work done on it over 50m?

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Answer: the acceleration of the object is 4 m/s^2.

Explanation: To determine the acceleration of the object, we can use the work-energy principle, which states that the work done on an object is equal to its change in kinetic energy. The equation for the work-energy principle is:

W = ΔKE = (1/2)mv^2 - (1/2)mu^2

where W is the work done, ΔKE is the change in kinetic energy, m is the mass of the object, v is the final velocity of the object, and u is the initial velocity of the object (which we assume to be zero).

We can rearrange this equation to solve for the final velocity v:

v^2 = (2W/m) + u^2

Since the initial velocity is zero, this simplifies to:

v^2 = 2W/m

Now, we can use the equation for average acceleration:

a = (v - u) / t

where t is the time taken to travel the distance of 50m. Assuming that the object starts from rest, u = 0, and the equation simplifies to:

a = v / t

Substituting the expression for v, we get:

a = sqrt(2W/m) / t

Plugging in the given values of W = 400 J, m = 2 kg, and t = 50 m / v (since t = d/v), we get:

a = sqrt(2*400 J / 2 kg) / (50 m / v)

a = sqrt(400 m^2/s^2) / (50 m / v)

a = 4 m/s^2

Therefore, the acceleration of the object is 4 m/s^2.

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