- Mass of the cart, \(m = 70 \text{ kg}\)

- Distance moved along the incline, \(d = 50 \text{ m}\)

- Angle of the incline, \(\theta = 45^\circ\)

- Coefficient of kinetic friction, \(\mu_k = 0\) (frictionless incline)

To find:

- Work done on the cart, \(W\)

Solution:

The work done on the cart is given by:

$$W = Fd\cos\theta$$

Since the incline is frictionless, the only force acting on the cart is the force of gravity parallel to the incline. This force is given by:

$$F = mg\sin\theta$$

Substituting this into the expression for work, we get:

$$W = mgd\sin\theta$$

Plugging in the given values, we get:

$$W = (70 \text{ kg})(9.8 \text{ m/s}^2)(50 \text{ m})\sin45^\circ$$

$$W = 15680 \text{ J}$$

Therefore, the work done on the cart is 15680 J.