a 2.0 kg particle is given a displacement (delta)r= 2.0i+3.0j (m). During this displacement, a constant force, F=2.0i-5j (N) acts at the particle. find the work done by the force for this displacement.
Work is the dot product of the Force and Displacement vectors.
I assume you know what a dot product is. If not, you need to review that subject.
The displacement of a bicycle S in a time t is given by S=A,+Bt+Ctsquer deduced the constants A,B,C appearing in the equation
To find the work done by a force, you can use the formula:
Work = Force • Displacement • cos(θ)
where Force is the magnitude of the force, Displacement is the displacement vector, and θ is the angle between the force and displacement vectors.
In this case, the force is given as F = 2.0i - 5.0j N and the displacement is given as Δr = 2.0i + 3.0j m.
First, let's find the magnitude of the force:
|F| = √(2.0^2 + (-5.0)^2)
= √(4.0 + 25.0)
= √29.0
≈ 5.39 N (rounded to two decimal places)
Next, let's find the dot product of the force and displacement vectors. The dot product is given by the formula:
Force • Displacement = Fx * Δrx + Fy * Δry
Using the given values:
Force • Displacement = (2.0 * 2.0) + (-5.0 * 3.0)
= 4.0 - 15.0
= -11.0
Finally, to find the angle between the force and displacement vectors, you can use the formula:
cos(θ) = (Force • Displacement) / (|F| * |Δr|)
Substituting the values:
cos(θ) = (-11.0) / (5.39 * √(2.0^2 + 3.0^2))
= (-11.0) / (5.39 * √(4.0 + 9.0))
= (-11.0) / (5.39 * √(13.0))
= (-11.0) / (5.39 * 3.61)
≈ -0.57 (rounded to two decimal places)
Now, we can calculate the work done:
Work = |F| * |Δr| * cos(θ)
= 5.39 * √(2.0^2 + 3.0^2) * (-0.57)
= 5.39 * √(4.0 + 9.0) * (-0.57)
= 5.39 * √(13.0) * (-0.57)
≈ -20.52 J (rounded to two decimal places)
So, the work done by the force during this displacement is approximately -20.52 J. The negative sign indicates that work is done against the force.