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A cyclist and her bicycle (combined mass of 75.0 kg) apply a force of 34.2 N. If the frictional force is -23.5 N, what will be the acceleration of the bicycle (in m/s2)? not sure about this question m: 75.0 kg F = 34.2 N Ff: -23.5 N a: ? F = m * a 34.2 N = 75.0 kg * a a = 34.2 N Well, what is the net force on the bicycle, if cyclist plus bicycle exert force of 34.2 N and frictional force opposes to the tune of 23.5 N? From that net force, work out acceleration ΣF = 34.2 N - 23.5 N ΣF = 10.7 N F = m * a a = F / m a = 10.7 N / 75.0 kg a = 0.14266 m/s^2 a = 0.143 m/s^2
yup. that looks correct. i got the same thing.
@nikato, well it was marked wrong... so i am trying to figure out what i did wrong...
ok... i'll try to figure it out what went wrong too
is there any more info for this problem?
@nikato no there is not
hi! @nikato and @Summersnow8 i took a look at what you both did and it looks ok to me. the only thing i can come up with to give this problem a twist is considering that the frictional force is applied to both wheels, so there should be 2 terms for friction. but taking a look at the numbers that wouldn't make much sense, so iguess i am lost too :(
Hi all! This is a very cool question. I think that the 34.2N thing should not be seen in the working as it is an internal force. The coolest thing is that the frictional force is to be added in the total external force and will result in the motion of the system(girl+bike). Always remember, the frictional force does not oppose the motion of an object, it only opposes the relative motion between the point of contacts of the objects. Imagine the point of contact of the rear wheel of the bicycle to be red in color. now assume that the wheel is in the air and can rotate freely. Now imagine in 20x slow motion(lol). The point of contact (in red) will appear to slowly go in the backward direction if the wheel is rotating in anticlockwise direction. Similarly with the rotating front wheel. Now, bring back the bike on the ground. So, the relative motion of the point of contact of both the wheels with respect to ground is in the backward direction so the frictional force will oppose the backward motion of the point of contact and the friction will be in the forward direction and will be a reason for the motion of the system. So, 34.2N and the frictional force are external forces for the bike. So, both the forces should be added in order to get the total force on the bike. so total force= 34.2N+23.5N=57.7N so acceleration= a = 57.7/75=0.76933333.... Please let me know if i'm correct or not. Because, if i'm not, i have another answer ready(lol). Hope this helps you!
@rajat97, the correct answer was a = 0.143 m/s^2
oops! I put too much of brains in it to get it wrong!(lol) i'm sorry!
@rajat97, it is okay
so you were right!