 Solution

ID: past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/section-b/6/c/solution

Let

m_{p} = 0.250 kg

and let

m_{w}

be the hanging mass. Taking up-slope and downward respectively as positive,

T - m_{p} g sin θ - μ m_{p} g cos θ = m_{p} a, m_{w} g - T = m_{w} a .

Eliminate the common tension to obtain

m_{w} = \frac{m _{p} { a + g ( sin θ + μ cos θ )}}{g - a} = 0.130 kg .

Here

θ = 1 0^{\circ}

μ = 0.1

and

a = 1.60 m s^{- 2}

. The pulley changes the direction of the string force; in the ideal model it does not change its magnitude.

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