Question

q2. a student pushes a box with a 50n force over a 3m distance along the floor. this force make an angle of 30° with the horizontal find the work done by the student
q3 a 10kg cart is pushed so that its velocity increases from 2m/s to 5m/s. using work - energy theorem, calculate the net work done on the cart.

Explanation:

Step1: Recall the work formula

The formula for work \(W = Fd\cos\theta\), where \(F = 50N\), \(d=3m\), \(\theta = 30^{\circ}\), and \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\approx0.866\).

Step2: Substitute values into the formula

\(W=(50N)\times(3m)\times\cos30^{\circ}\)
\(W = 50\times3\times0.866\)
\(W=150\times0.866 = 129.9J\approx130J\)

for Q3:

Step1: Recall the work - energy theorem

The work - energy theorem states that \(W_{net}=\Delta K.E\), and the kinetic energy formula \(K.E=\frac{1}{2}mv^{2}\). Here, \(m = 10kg\), \(v_{1}=2m/s\), \(v_{2}=5m/s\)

Step2: Calculate the change in kinetic energy

\(\Delta K.E=\frac{1}{2}m(v_{2}^{2}-v_{1}^{2})\)
Substitute \(m = 10kg\), \(v_{1}=2m/s\), \(v_{2}=5m/s\)
\(\Delta K.E=\frac{1}{2}\times10\times(5^{2}-2^{2})\)
\(= 5\times(25 - 4)\)
\(=5\times21\)
\(=105J\)

Answer:

The work done by the student is approximately \(130J\)