Hey. I'm wondering if I have answered the following questions correctly.

Question:

Anna first ride a bike (case 1) on a 100 m horizontal distance and then (case 2) she rides up a 50m long slope with a height difference of 7m. Anna weighs 54 kg and her bike weighs 16 kg.

a. How much work is done vertically in case 1? (Neglecting any impact of friction)

b. How much work is done vertically in case 2? (Neglecting friction)

c. What force does the work in question b) ?

Solution:

a. By Newtons second law the vertical resultant force \(\displaystyle F_{y}=0 \iff W_y = F_{y}(100m) =0 \). Since the verticall resultant force is zero no work is performed vertically.

b.

Assume she is travelling with constant velocity. Then the force of gravity which is acting on her and the bike is equal to \(\displaystyle Fg =mg \). Denote the incline angle \(\displaystyle \alpha \). And define the \(\displaystyle {\parallel}\) -axis parallel to the contact surface and the

\(\displaystyle \perp \) axis perpindicular to \(\displaystyle {\parallel}\) -axis. The force of gravity can be broken into its component forces. Now the resultant force \(\displaystyle F_{\perp} =0\) and \(\displaystyle F_{\parallel}=0\) since she is travellling with constant velocity. Denote the force of gravity with is parellel to the contact surface \(\displaystyle Fg_{\parallel}\). By trigonomety it follows that \(\displaystyle Fg_{\parallel}=Sin \alpha \space Fg= Sin \alpha \space mg\). And by newtons second law if the resultant force \(\displaystyle F_{\parallel}=0\) then there exists a force \(\displaystyle Fg^*_{\parallel}\) such that \(\displaystyle Fg_{\parallel}+Fg^{*}_{\parallel}=0\). Then the total amount of work performed while she is travelling up the incline is equal to \(\displaystyle W_{total}=Fg^{*}_{\parallel}(50m)=- Sin \alpha \space mg (50m) \) .

We know that \(\displaystyle g = -9,82 m/s^2\), \(\displaystyle \alpha = Sin^{-1}(7/50) \approx 8,0478 ^{\circ} \) and \(\displaystyle m = 70 kg\). Hence it follows that \(\displaystyle W_{total} \approx 4,8 kJ\).

One can also interpret the question differently. Assume by horizontal they mean the distance she travelled from case 1. Now the force which is vertical to the horizontal line is just \(\displaystyle Fg\). And since the vertical resultant force is zero there must exist an equal and opposite acting force call it \(\displaystyle Fg^{*}=-Fg\). And the work performed in the vertical direction \(\displaystyle W_{vertical} = -Fg(7m)=-mg(7m)\approx 4,8 kJ\). In either case, we obtain the same value.

c.

The forces \(\displaystyle Fg^{*}_{\parallel}\) and \(\displaystyle Fg^{*}\) is the exact amount of force necessary to oppose the equal and opposite directed force of gravity to maintain a constant velocity during the travelling up the incline.

Question:

Anna first ride a bike (case 1) on a 100 m horizontal distance and then (case 2) she rides up a 50m long slope with a height difference of 7m. Anna weighs 54 kg and her bike weighs 16 kg.

a. How much work is done vertically in case 1? (Neglecting any impact of friction)

b. How much work is done vertically in case 2? (Neglecting friction)

c. What force does the work in question b) ?

Solution:

a. By Newtons second law the vertical resultant force \(\displaystyle F_{y}=0 \iff W_y = F_{y}(100m) =0 \). Since the verticall resultant force is zero no work is performed vertically.

b.

Assume she is travelling with constant velocity. Then the force of gravity which is acting on her and the bike is equal to \(\displaystyle Fg =mg \). Denote the incline angle \(\displaystyle \alpha \). And define the \(\displaystyle {\parallel}\) -axis parallel to the contact surface and the

\(\displaystyle \perp \) axis perpindicular to \(\displaystyle {\parallel}\) -axis. The force of gravity can be broken into its component forces. Now the resultant force \(\displaystyle F_{\perp} =0\) and \(\displaystyle F_{\parallel}=0\) since she is travellling with constant velocity. Denote the force of gravity with is parellel to the contact surface \(\displaystyle Fg_{\parallel}\). By trigonomety it follows that \(\displaystyle Fg_{\parallel}=Sin \alpha \space Fg= Sin \alpha \space mg\). And by newtons second law if the resultant force \(\displaystyle F_{\parallel}=0\) then there exists a force \(\displaystyle Fg^*_{\parallel}\) such that \(\displaystyle Fg_{\parallel}+Fg^{*}_{\parallel}=0\). Then the total amount of work performed while she is travelling up the incline is equal to \(\displaystyle W_{total}=Fg^{*}_{\parallel}(50m)=- Sin \alpha \space mg (50m) \) .

We know that \(\displaystyle g = -9,82 m/s^2\), \(\displaystyle \alpha = Sin^{-1}(7/50) \approx 8,0478 ^{\circ} \) and \(\displaystyle m = 70 kg\). Hence it follows that \(\displaystyle W_{total} \approx 4,8 kJ\).

One can also interpret the question differently. Assume by horizontal they mean the distance she travelled from case 1. Now the force which is vertical to the horizontal line is just \(\displaystyle Fg\). And since the vertical resultant force is zero there must exist an equal and opposite acting force call it \(\displaystyle Fg^{*}=-Fg\). And the work performed in the vertical direction \(\displaystyle W_{vertical} = -Fg(7m)=-mg(7m)\approx 4,8 kJ\). In either case, we obtain the same value.

c.

The forces \(\displaystyle Fg^{*}_{\parallel}\) and \(\displaystyle Fg^{*}\) is the exact amount of force necessary to oppose the equal and opposite directed force of gravity to maintain a constant velocity during the travelling up the incline.

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