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ssm A bicyclist is coasting straight down a hill at a constant speed. The combined mass of the rider and bicycle is 80.0 $\mathrm{kg}$ , and the hill is inclined at $15.0^{\circ}$ with respect to the horizontal. Air resistance opposes the motion of the cyclist. Later, the bicyclist climbs the same hill at the same constant speed. How much force (directed parallel to the hill) must be applied to the bicycle in order for the bicyclist to climb the hill?

406 $\mathrm{N}$

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Cornell University

University of Michigan - Ann Arbor

Numerade Educator

University of Washington

in both situations. We're going to use the following reference frame. This axis will be the why access and these will be the X axis. Now we can apply Newton's second law on both directions. So far. Situation number one Applying Newton's second law on the Y axis gives us the following. The Net force in this direction is given by the maths off the writer and the bicycle times their acceleration in that direction. Note that the bike is not going to move a ports like this or downwards like this. Then it's acceleration in that direction is a question zero. So the net force in the Y direction is It goes to zero and the net force is composed by two forces in the direction we have the normal force, which is pointing to the positive Y direction. So normal force and we also have a component off the weight force, which is pointing to the negative y direction. Let me call you w component. Why? So the normal force minors the weight component. Why shortly equals 20 Then the normal force is you close to the y component off the weight force? No, for the acts access. We have the following on the X direction than that force is equals to the combine it mass times acceleration, index direction. Note that the situation the bicycle is going downhill with a constant velocity, so its acceleration is close to zero index direction, then the net forest next direction is it close to zero. But the Net force index direction is composed by two forces. The air resistance which points to the negative X direction, and the X component off the weight force which points to the positive X direction. Then the wait for its component acts minus the air resistant force is a close to zero. So you the Air Resistance Force is in question. The weight component acts now in situation number two. We are interested in calculating this force, which is the apply it force that is neccessary for the bicycle to go up. He'll then we should apply Newton's second law here first X direction. So the net force next direction is. It goes to the mass of the bicycle times its acceleration in the X direction again, it's going up Hugh with a constant velocity. So its acceleration in the X direction is it close to zero. Then the net force next direction is again equals zero. Now the net force in next directions composed by three forces the apply it force the air resistance forge and the X component off the weight force which we can draw as follows. So these will be the X component off the weight force and these with me it's why component, then the net force next direction can be written as follows. So pointing to the positive X direction we have the Air Resistance Force and the weight component X and then pointing to the negative X direction we have. The applied force on these wolfing should be equal to zero. Then they apply a force should be equal to the at resistance force plus the weight force component X But note that from situation number one we discovered that the weight component acts is equals to the air resistance Force. Then they apply it Force should be equals to the wait component X plus the weight component X then the applying force is equals to two times the weight component acts. Now how can he calculates the X component off the weight force? We have to take a look at this triangle. So we have 90 degrees in here and 50 degrees there. So why does it see if two degrees? Well, you can note that this angle here should be. If this is 15 degrees, it should be 75 degrees. Now we can extend the weight force, and this makes a 90 degree angle with the ground, and then these is a triangle so its internal angles shoot. Add up to 180 and we have 90 year 75 here and 15 here 75 plus 15 is equal to 90 then. In fact, this triangle has intern angles that add up to 180 before this angle is really 15 degrees. Then we can use the sign off 15 degrees to calculate the wave component. X, because off the sign off, 15 degrees in this context will be given by the opposite side of triangle, which is the weight component acts divided by the partners, which is the weight. Then the weight component acts. Is it close to the weight kinds? The sign off 15 degrees to the applying force. She'll be equals two, two times the wait times this sign off 15 degrees. But the weight force remember is given by the mass times acceleration of gravity, which is approximately 9.8 meters per second squared, and the mass. The combined mass is 80 kilograms. Therefore apply IT force is given by two times 18 times 9.8 times this sign off 15 degrees, and these results in an apply IT force off approximately 406 new tones.

Brazilian Center for Research in Physics