Answered

Westonci.ca makes finding answers easy, with a community of experts ready to provide you with the information you seek. Explore our Q&A platform to find reliable answers from a wide range of experts in different fields. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

A 5 .0kg cart is moving horizontaly at 6 .0m/s in order to change its speed to 10m/s the net work done on the cart must be

Sagot :

onyebuchinnaji

Answer:

the net work done on the cart is 160 J.

Explanation:

Given;

mass of the cart, m = 5.0 kg

initial velocity of the cart, u = 6 m/s

final velocity of the cart, v = 10 m/s

The net work done on the cart is equal to change in average kinetic energy of the cart;

[tex]W = K.E = \frac{1}{2} m(v^2-u^2)\\\\W = \frac{1}{2} \times 5(10^2-6^2)\\\\W = 160 \ J[/tex]

Therefore, the net work done on the cart is 160 J.

Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.