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Problem 115 Hard Difficulty

ssm Refer to Multiple-Concept Example 10 for help in solving problems like this one. An ice skater is gliding horizontally across the ice with an initial velocity of $+6.3 \mathrm{m} / \mathrm{s}$. The coefficient of kinetic friction between the ice and the skate blades is 0.081, and air resistance is negligible. How much time elapses before her velocity is reduced to $+2.8 \mathrm{m} / \mathrm{s} ?$


4.4 $\mathrm{s}$


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Video Transcript

to solve this question, we begin by calculating the acceleration off the skater. For that, we have to use Newton's second law, and I'm choosing the following reference frame vertical Y axis on the horizontal X axis. Then applying Newton's second law to my vertical axes results in the following the Net force. That direction is equal to the mass off the skater times its vertical acceleration. Notice that it's not moving in the vertical direction and it's not going to move in that direction. Therefore, the acceleration is close to zero and the net force is it close to zero. But the net force in the Y direction is composed by the normal force and the weight force. The normal force points, the positive direction and the wait for the negative direction. Then we conclude that the normal force is equal to the weight force office cater, which is given by his mass times. The acceleration of gravity nearness surface off the earth. Repeating this, Bruce says, For the X axis, we get that the net force acting the X direction is given by his mass times, his ex acceleration, then the net force acting on the horizontal axis is given by the frictional force only which points the negative direction so minus frictional force is equals to his mass times. His acceleration in the X direction but the frictional force in the situation where he's already moving is given by the kinetic fictional quest. He shan't times the normal force. So miners in that exclusion, all coefficients times the normal force is equals to his mass times his ex acceleration. And you saw that the normal force is equal to the weight force which is equals to sometimes g. Then, using these results here, we got the following miners from UK times, M times G is equals to m times. He's ex acceleration. We can simplify the masses and get on expression for his acceleration, which is minus mu k times G. The acceleration of gravity near the surface off the earth is approximately 9.8 meters per second squared. Then his acts acceleration is minus 0.0 81 times 9.8, which results in an acceleration off minus 0.7938 meters per second squared. This is his axe acceleration. Now we can calculate how much time elapses before his velocity is reduce it to true 0.8 meters per second. For that, remember that the velocity as a function off time is given by the initial velocity plus the acceleration times the time interval This question. We have a final velocity off true 0.8 meters per second and an initial velocity off 6.3. The accelerations minor 0.7938 later Wolf time is not now. Then we can solve this equation for time to get our answer. This goes as follows. We begin by sending this term to the other side in these terms to the other side. So 0.7938 times cheek is equals to 6.3 minus 2.8 them tea his e coast of 6.3 miners 2.8 divided by 0.7938 and these results in an interval of time off approximately four 0.4 seconds. These is down. Started this problem