A box with a mass of 7 kilograms is pushed up a ramp to a height of 5 meters. What is the work done against the force of gravity
2 answers:
The correct answer to the question is .
CALCULATION:
As per the question, the mass of the box is given as m = 7 kg.
The box was pushed to a height h = 5 m.
We are asked to calculate the work done by the box.
We know that energy is the capacity to do work.The work done by a body when it is taken to certain height above the surface of earth against gravity is equal to the potential energy gained by that body .
Hence, the work done W = P.E
The potential energy gained by the box when it was taken to certain height is the gravitational potential energy of the box.
The gravitational potential energy gained by the box is calculated as -
Potential energy P.E = mgh
= 7 × 9.8 × 5 J.
= 343 J
=
=
Here, g is known as the acceleration due to gravity and g =
We know that work done by the box against gravity W = P.E.
Hence, the work done by the box against gravity is
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Hi there!
a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.
dB = Differential Magnetic field element
μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)
R = radius of loop (2.15 cm = 0.0215 m)
i = Current in loop (0.460 A)
For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.
Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
Taking out constants from the integral:
Since we are integrating around an entire circle, we are integrating from 0 to 2π.
Evaluate:
Plugging in our givens to solve for the magnetic field strength of one loop:
Multiply by the number of loops to find the total magnetic field:
b)
Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.
Using the diagram, if 'z' is the point's height from the center:
Substituting this into our expression:
Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
Evaluate:
Multiplying by the number of loops:
Plug in the given values: