HELP ME, PLEASE.

At the bottom of a frictionless ramp, a 2kg mass reaches a speed of 9 m/s, how high is the top of the ramp? *

noname [10]

V=√(2gs)
9=√(2×9.8×s)
s=4.1m

3 0

3 years ago

The 4.0 -kg head of an ax is moving at 3 m/s when it strikes a log and penetrates 0.01m into the log. What is the average force

Elina [12.6K]

Answer:

the average force the blade exerts on the log is 1791.05 N.

Explanation:

Given;

mass of the ax head, m = 4 kg

speed of the ax, v = 3 m/s

depth traveled into the log, d = 0.01 m

The time to traveled through the depth;

$s = (\frac{u+v}{2} )t\\\\0.01 = (\frac{0+3}{2} )t\\\\0.01 = 1.5t\\\\t = \frac{0.01}{1.5} \\\\t = 0.0067 \ s$

The average force the blade exerts on the log;

$F = ma=\frac{mv}{t} = \frac{4 \times 3}{0.0067} = 1791.05 \ N \\\\$

Therefore, the average force the blade exerts on the log is 1791.05 N.

5 0

3 years ago

A body of mass 200gram is executing SHM with amplitude of 20mm. The magnitude of maximum force which acts upon it is 0.6N. Calcu

igor_vitrenko [27]

The value of maximum velocity will be 0.3464 m/s². Energy or work is equal to the product of force and displacement.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

Mass of the body is,(m= 200g)

Amplitude is,(A =20 mm)

Maximum force is,F= 0.6 N

To find;

Maximum velocity

Energy or work is equal to the product of force and displacement.

$\rm KE = \frac{1}{2} mV_{max}^2 \\\\ \rm F_{max} \times A = \frac{1}{2} mV_{max}^2 \\\\ F_{max}= \frac{1}{2}\frac{ mV_{max}^2}{A} \\\\ 0.6= \frac{1}{2}\times \frac{ 0.2 \times V_{max}^2}{20 \times 10^{-3}} \\\\ V_{max}=0.3464 \ m/sec^2$