UQAM PHY0350 tutoring

1. Resolve forces along the incline, choosing downhill as positive. Gravity’s component in that direction is $mg\sin\theta$. The contact force exerted by the incline, called the normal force, is perpendicular to the incline. It therefore has no component along the motion, and there is no friction. With $a$ denoting the downhill acceleration, Newton’s law gives $ma=mg\sin\theta$, hence $a=g\sin30^\circ=4.9\,\mathrm{m/s^2}$. 2. Use the constant acceleration. Let $v_0=0$ be the initial speed and $v$ the final speed. Then $v^2=v_0^2+2as=0+2(4.9\,\mathrm{m/s^2})(1.25\,\mathrm{m})=12.25\,\mathrm{m^2/s^2}$. Speed is nonnegative, so $v=3.5\,\mathrm{m/s}$. 3. Check with energy. Let $h$ be the vertical drop; geometry gives $h=s\sin\theta=1.25\sin30^\circ=0.625\,\mathrm{m}$. The lost gravitational potential energy is $mgh=(2)(9.8)(0.625)=12.25\,\mathrm{J}$. It equals the gained kinetic energy $\frac12mv^2$, giving the same speed. The normal force does no work because it is perpendicular to the displacement. 4. Explain the mass dependence. In the force equation the mass cancels, and in the energy equation $mgh=\frac12mv^2$ it cancels again, leaving $v=\sqrt{2gh}$. Doubling the mass therefore leaves the speed unchanged under these assumptions. Using the distance along the incline as the vertical drop would overestimate the available energy.