A 0.260 kg potato is tied to a string with length 2.50 m, and the other end of the string is tied to a rigid support. The potato is held straight out horizontally from the point of support, with the string pulled taut, and is then released. A) What is the speed of the potato at the lowest point of its motion? let g be = 9.80 m/s^2
B) What is the tension on the string at this point?

answers

velocity = 7 m/s

tension = 7.644  N

set up equation for velocity

because energy is conserved, the change in potential energy as the potato drops is the same as the change in kinetic energy

U = K

mgh = m/2

gh = /2

= 2gh

v =

values

g = 9.8 m/s^2

h = 2.5 m (change in vertical distance)

plug in values and solve

v =

v =

v = 7 m/s

set up equation for tension

at the lowest point, tension (directly upward) acts in the opposite direction of weight (directly downwards) so

∑F = T - mg

since the potato is in circular motion, the net force is equal to centripetal force

∑F = m/r

∑F = T - mg = m/r

T - mg = m/r

T = mg + m/r

T =

values

m = 0.26 kg

g = 9.8 m/s^2

v = 7 m/s

r = 2.5m

plug in values and solve

T =

T =

T = 7.644  N

d. air pressure decreases as altitude increases.

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft