What is the mass of the heaviest book this person can hold onto vertically before it slips out of his or her fingers? The coeffi
1 answer:
Answer:
0.795 kg
Explanation:
Assuming the complete question:
A person with compromised pinch strength in their fingers can only exert a normal force of 6.0 N to either side of a pinch-held object. What is the mass of the heaviest book this person can hold onto vertically before it slips out of his or her fingers? The coefficient of static friction of the surface between the fingers and the book cover is 0.60
SOLUTION:
The maximum weight of the book will equal the maximal friction force that can be produced:
m g = 2 f
Note that there are two sides of the book, so the friction force equals 2 times the friction force on one side (hence the factor 2).
So the maximum mass of the book is
m = 2 f / g
m = 20.65
6.0N / (9.81N/kg)
m = 0.795 kg
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Answer:
(a) 348.4 m
(b) 256.7 m/s
(c) 127.2 m/s^2
Explanation:
(a) at t = 4 s
x = 2.3 x 4 + 5.3 x 4 x 4 x 4
x = 348.4 m
(b) The derivative of displacement function gives the value of instantaneous velocity.
So, v = dx / dt = 2.3 + 5.3 x 3 x t^2
v = 2.3 + 15.9 t^2
Put t = 4 s
So, v = 2.3 + 15.9 x 4 x 4
v = 256.7 m/s
(c) The derivative of velocity function with respect to time gives the value of instantaneous acceleration.
So, a = dv / dt = 5.3 x 3 x 2 x t
a = 31.8 t
Put t = 4 s
a = 31.8 x 4 = 127.2 m/s^2