A 5kg sled is sliding down a 20degree hill. There is friction.
a) how strong is the gravitational pull on the sled
b) how strong is the hill pushing (perpendicularly) on the sled? (I got this using mgcos(theta) and got 46.04N, is this correct?)
c) what is the net force from these two force vectors?
d) what is the theoretical acceleration of the sled?
e) if the force of friction is 10N, what is the new net force?
f) what is the actual acceleration of the sled downhill?
g) find mu of kinetic force for the sled along the hill.
I have been trying to figure out how to solve these problems for a half hour already, so honestly any help would be very much appreciated! Thank you!
a) how strong is the gravitational pull on the sled
b) how strong is the hill pushing (perpendicularly) on the sled? (I got this using mgcos(theta) and got 46.04N, is this correct?)
c) what is the net force from these two force vectors?
d) what is the theoretical acceleration of the sled?
e) if the force of friction is 10N, what is the new net force?
f) what is the actual acceleration of the sled downhill?
g) find mu of kinetic force for the sled along the hill.
I have been trying to figure out how to solve these problems for a half hour already, so honestly any help would be very much appreciated! Thank you!
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a) how strong is the gravitational pull on the sled: 5*9.81= 49.05 N vertically downwards
b) how strong is the hill pushing (perpendicularly) on the sled? (I got this using mgcos(theta) and got 46.04N, is this correct?): You are right (49.05 *cos 20 = 46.09 N)
c) what is the net force from these two force vectors? = 46.09 N gets balanced. 49.05*sin 20 = 16.78 N down the sled remains
d) what is the theoretical acceleration of the sled?(By theoretical you mean excluding friction, 16.78 N /5 = 9.81 *sin 20 = 3.355 m/s²
e) if the force of friction is 10N, what is the new net force?: 16.78 - 10 = 6.78 N
f) what is the actual acceleration of the sled downhill? 6.78/5 = 1.355 m/s²
g) find mu of kinetic force for the sled along the hill. 10/46.09 = 0.2169 or 0.217
b) how strong is the hill pushing (perpendicularly) on the sled? (I got this using mgcos(theta) and got 46.04N, is this correct?): You are right (49.05 *cos 20 = 46.09 N)
c) what is the net force from these two force vectors? = 46.09 N gets balanced. 49.05*sin 20 = 16.78 N down the sled remains
d) what is the theoretical acceleration of the sled?(By theoretical you mean excluding friction, 16.78 N /5 = 9.81 *sin 20 = 3.355 m/s²
e) if the force of friction is 10N, what is the new net force?: 16.78 - 10 = 6.78 N
f) what is the actual acceleration of the sled downhill? 6.78/5 = 1.355 m/s²
g) find mu of kinetic force for the sled along the hill. 10/46.09 = 0.2169 or 0.217