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force and acceleration 1. skelly the skater, total mass 25 kg, is prope…

Question

force and acceleration

  1. skelly the skater, total mass 25 kg, is propelled by rocket power.

a. complete table i
(neglect resistance)
table i
b. complete table ii for a
constant 50 - n resistance.
table ii

  1. block a on a horizontal friction - free table is accelerated by a force from

a string attached to block b. b falls vertically and drags a horizontally.
both blocks have the same mass m. (neglect the strings mass.)
circle the correct answers.
a. the mass of the system (a + b) is (m) (2m).
b. the force that accelerates (a + b) is the weight of (a) (b) (a + b).
c. the weight of b is (mg/2) (mg) (2mg).
d. acceleration of (a + b) is (less than g) (g) (more than g).
if b were allowed to fall by itself, not dragging a,
then wouldnt its acceleration be g?
yes, because the force that accelerates
it would only be acting on its own
mass - not twice the mass!
to better understand this,
consider 3 and 4 on the
other side!

Explanation:

Step1: Calculate acceleration for Table I (using \(F = ma\), \(a=\frac{F}{m}\))

For \(F = 100N\), \(m = 25kg\), \(a=\frac{100}{25}=4m/s^{2}\)
For \(F = 200N\), \(m = 25kg\), \(a=\frac{200}{25}=8m/s^{2}\)
For \(a = 10m/s^{2}\), \(m = 25kg\), \(F=ma = 25\times10 = 250N\)

Step2: Calculate acceleration for Table II (using \(F_{net}=F - F_{res}\), \(a=\frac{F_{net}}{m}\))

For \(F = 50N\), \(F_{res}=50N\), \(F_{net}=50 - 50=0N\), \(a = 0m/s^{2}\) (already given)
For \(F = 100N\), \(F_{res}=50N\), \(F_{net}=100 - 50 = 50N\), \(a=\frac{50}{25}=2m/s^{2}\)
For \(F = 200N\), \(F_{res}=50N\), \(F_{net}=200 - 50=150N\), \(a=\frac{150}{25}=6m/s^{2}\)

Step3: Answer sub - questions for Block A and B

a. The mass of system \((A + B)\) is \(m + m=2m\)
b. The force that accelerates \((A + B)\) is the weight of \(B\) (since \(B\) is the part of the system providing the driving force due to gravity)
c. Weight of \(B\) is \(W = mg\) (by formula \(W=mg\))
d. Using \(F = ma\), \(F = mg\) (weight of \(B\)), \(m_{total}=2m\), \(a=\frac{mg}{2m}=\frac{g}{2}\), so acceleration of \((A + B)\) is less than \(g\)

Answer:

Table I:

FORCEACCELERATION
\(200N\)\(8m/s^{2}\)
\(250N\)\(10m/s^{2}\)

Table II:

FORCEACCELERATION
\(100N\)\(2m/s^{2}\)
\(200N\)\(6m/s^{2}\)

a. \(2m\)
b. \(B\)
c. \(mg\)
d. less than \(g\)