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June 2005 p4 q1
3618
A small block is pulled along a rough horizontal floor at a constant speed of 1.5 m s-1 by a constant force of magnitude 30 N acting at an angle of \(\theta^\circ\) upwards from the horizontal. Given that the work done by the force in 20 s is 720 J, calculate the value of \(\theta\).
Solution
The work done \((WD)\) by the force is given by the formula:
\(WD = Fd \cos \theta\)
where \(F = 30 \text{ N}\) is the force, \(d\) is the distance moved, and \(\theta\) is the angle of the force above the horizontal.
Since the block moves at a constant speed of 1.5 m/s for 20 seconds, the distance \(d\) is:
\(d = 1.5 \times 20 = 30 \text{ m}\)
Substituting the given values into the work done formula: