A box of mass $$$5 \mathrm{~kg}$$$ is pulled at a constant speed a distance of $$$15 \mathrm{~m}$$$ up a rough plane inclined at an angle of $$$20^{\circ}$$$ to the horizontal. The box moves along a line of greatest slope against a frictional force of $$$40 \mathrm{~N}$$$. The force pulling the box is parallel to the line of greatest slope. Find the change in gravitational potential energy of the box. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mathematics

IGCSE&ALevel

CAIE

Exam No：9709_w20_qp_43 Year：2020 Question No：2(b)

Answer:

$\begin{array}{l}{[P E=5 \times 10 \times 15 \sin 20]} \\ 257 \mathrm{~J} \quad(256.5151 \ldots \mathrm{J})\end{array}$

Knowledge points:

4.5.1 understand the concept of the work done by a force, and calculate the work done by a constant force when its point of application undergoes a displacement not necessarily parallel to the force W = Fd cos; Use of the scalar product is not required.

Answer:

Knowledge points:

Solution:

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