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when there is friction, which mechanical energy - time diagram is correct?
potential energy
total energy
kinetic energy
Step1: Analyze potential energy
When an object slides down a slope with friction, its height decreases. According to the formula \(E_p = mgh\) (where \(m\) is mass, \(g\) is gravitational acceleration, \(h\) is height), potential energy \(E_p\) decreases. But it doesn't decrease linearly because the acceleration \(a=\frac{mg\sin\theta - f}{m}\) (\(f\) is frictional force, \(\theta\) is the slope angle), and \(h = h_0-\frac{1}{2}at^{2}\) (initial - \(h_0\)), so \(E_p=mgh=mgh_0-\frac{1}{2}mgat^{2}\), which is a quadratic - like relationship (not linear).
Step2: Analyze kinetic energy
From the work - energy theorem \(W_{net}=\Delta K\). The net work \(W_{net}=(mg\sin\theta - f)x\) (\(x\) is the displacement along the slope). And \(x = v_0t+\frac{1}{2}at^{2}\) (if \(v_0 = 0\), \(x=\frac{1}{2}at^{2}\)), so \(K=\frac{1}{2}(mg\sin\theta - f)at^{2}\), which is a quadratic function of time \(t\) (parabola opening upwards at the beginning).
Step3: Analyze total energy
Due to friction, the total mechanical energy \(E = E_p+K\) decreases. The frictional force \(f\) does negative work \(W_f=-fx\), and \(E = E_0+W_f\) (\(E_0\) is the initial total mechanical energy). Since \(x\) is related to \(t\) non - linearly (\(x=\frac{1}{2}at^{2}\) for \(v_0 = 0\)), the total energy decreases non - linearly.
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