Easy -1.2 This is a straightforward application of Newton's second law (F = ma) with a single object on a horizontal surface. Students need only identify forces, apply F_net = ma, and solve a simple linear equation: E - 400 = 6000 × 0.2, giving E = 1600 N. It's a standard textbook exercise requiring basic recall and arithmetic, making it easier than average.
\includegraphics{figure_1}
An engine pulls a truck of mass 6000 kg along a straight horizontal track, exerting a constant horizontal force of magnitude \(E\) newtons on the truck (see diagram). The resistance to motion of the truck has magnitude 400 N, and the acceleration of the truck is \(0.2 \text{ m s}^{-2}\). Find the value of \(E\). [4]
For resultant force \(E - 400\) stated or implied; For use of Newton II for the truck; For the correct equation; For correct answer 1600
Hence \(E = 1600\)
$E - 400 = 6000 \times 2$ | B1, M1, A1, A1 | For resultant force $E - 400$ stated or implied; For use of Newton II for the truck; For the correct equation; For correct answer 1600
Hence $E = 1600$ | | |
\includegraphics{figure_1}
An engine pulls a truck of mass 6000 kg along a straight horizontal track, exerting a constant horizontal force of magnitude $E$ newtons on the truck (see diagram). The resistance to motion of the truck has magnitude 400 N, and the acceleration of the truck is $0.2 \text{ m s}^{-2}$. Find the value of $E$. [4]
\hfill \mbox{\textit{OCR M1 Q1 [4]}}